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CBSE Questions for Class 12 Medical Physics Atoms Quiz 14 - MCQExams.com
CBSE
Class 12 Medical Physics
Atoms
Quiz 14
The ionization energy of hydrogen atom is $$13.6\ eV$$. Hydrogen atoms in the ground state are excited by electromagnetic radiation of energy $$12.1\ eV$$. How many spectral lines will be emitted by the hydrogen atoms?
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one
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two
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three
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four
Explanation
Energy of electron in hydrogen atom in ground state=-13.5eV.
When a photon of energy of 12.1eV is radiated on atom, the energy gained by the electron is given by :
$$E=(-13.6)+12.1=(-1.5)eV$$
And we know that energy of an excited electron is given by,
$$E=\dfrac { -13.6 }{ { n }^{ 2 } } (n:number\quad of\quad orbit)$$
$$\Rightarrow -1.5=\dfrac { -13.6 }{ { n }^{ 2 } } \\ \Rightarrow n=3$$
Therefore, when a photon of 12.6eV is radiated electron jump to 3 orbit.
Ionization potential of hydrogen is $$13.6$$ volt. If it is excited by a photon of energy $$12.1eV$$, then the number of lines in the emission spectrum will be:
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$$2$$
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$$3$$
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$$4$$
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$$5$$
Explanation
In this case, the energy of H-atom after it absorbs photon of energy 12.1eV will be $$=-13.6eV+12.1eV=-1.5eV$$
Since,
$$E_n=-R_H\left(\frac{1}{n^2}\right)$$
$$n^2=\frac{-R_H}{E_n}=\frac{-2.18\times 10^{-18}}{-1.5\times 1.6\times 10^{-19}}=9.08$$
$$n=3$$
The gyromagnetic ratio of an electron in an $$H-$$atom, according to Bohr model, is:-
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Independent of which orbit it is in
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Negative
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Positive
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Increase with the quantum number $$n$$
Explanation
$$\begin{array}{l}\text { we know that the - } \\\text { Gyromagnetic Ratio = } \frac{q}{2 m} \\\text { Since, the electron carries a negative charge. } \\\text { Hence, the gyromagnetic ratio will be negative .}\end{array}$$
There are some atoms in ground energy level and rest of all in some upper energy level in a gas of identical hydrogen like atoms. The atoms jump to higher energy level by absorbing monochromatic light of photon energy $$1.89\ eV$$ and this emit radiations of $$3$$ photons energies. Find the principal quantum number of initially excited level.
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$$3$$
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$$4$$
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$$2$$
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$$5$$
The figure shows the velocity and acceleration of a point like body at the initial moment of its motion. The acceleration vector of the body remains constant. The minimum radius of curvature of trajectory of the body is (in m)
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2 m
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4 m
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8 m
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16 m
According to Bohr's theory of hydrogen atom, the speed of the electron, its energy and the radius of its orbit varies with the principal quantum number $$n$$, respectively as:
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$$\cfrac { 1 }{ n } ,\cfrac { 1 }{ { n }^{ 2 } } ,{ n }^{ 2 }\quad $$
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$$\cfrac { 1 }{ n } ,{ n }^{ 2 },\cfrac { 1 }{ { n }^{ 2 } } $$
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$${ n }^{ 2 },\cfrac { 1 }{ { n }^{ 2 } } ,{ n }^{ 2 }$$
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$$n,\cfrac { 1 }{ { n }^{ 2 } } ,\cfrac { 1 }{ { n }^{ 2 } } $$
Taking the Bohr radius as $$a_{0}=35pm$$, the radius of $$Li^{++}$$ion in its ground state, on the basis of Bohr's model, will be about
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$$53pm$$
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$$27\ pm$$
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$$18\ pm$$
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$$13\ pm$$
Explanation
$$\begin{array}{l}\text { Given, } \\\text { Bohr's radius }=53 \mathrm{pm} \\\text { we know, } \\\text { Bohr's radius }=\frac{0.529 n^{2}}{2}\end{array}$$
$$\begin{array}{l}\text { if } n=1 & Z=1 \text { , the radius obtained is bohr's radius} \\\end{array}$$
$$\begin{array}{l}\text { radius }\left(a_{0}\right)=\frac{0.529(1)^{2}}{1} A=53\mathrm{Pm} \\\text { for } Li^{++} \quad n=1 \quad(\text { ground state }) \\\text { and } z=3\end{array}$$
$$\begin{aligned}\text { radius } &=\frac{53\cdot n^{2}}{2} p_{m}=\frac{53(1)^{2}}{3} \\r &=17.6 \approx 18 \mathrm{pm} \\\text { hence option (c) is wrrect. }\end{aligned}$$
Consider Bohr's theory for hydrogen atom. The magnitude of angular momentum, orbit radius and frequency of the electron in $$n^th$$ energy state in a hydrogen atom are, $$r$$ and $$f$$ respectively. find out the value of $$X$$, if ($$frl$$) is directly proportional to $$n^x$$.
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1
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0
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2
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3
Explanation
We know that-
$$r\propto n^2$$
$$v\propto \dfrac{1}{n}$$
and $$l=\dfrac{nh}{2\pi}\implies l\propto n$$
$$f=\dfrac{1}{T}=\dfrac{v}{2\pi r}$$
$$\implies f\propto \dfrac{1}{n^3}$$
Hence, $$frl \propto n^0$$
Hence, $$x=0$$
Electron in a hydrogen atom is replaced by an identically charged particle muon with mass $$207$$ times that of electron . Now the radius of K shell will be
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$$2.56 \times 10^{-3}\mathring {A}$$
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$$109.7 \mathring{A}$$
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$$1.21 \times 10^{-3}\mathring{A}$$
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$$22174.4 \mathring {A}$$
An ionic atom equivalent to hydrogen atom has wavelength equal to $$\dfrac{1}{4}$$ of the wavelengths of hydrogen lines. The ion will be
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$$He^+$$
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$$Li^{++}$$
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$$Ne^{9+}$$
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$$Na^{10+}$$
Explanation
$$\text{By Rydberg's formula ,we have -}$$
$$\begin{array}{l}R_{H}=\text { Rydberg constant } \\\lambda=\text { wavelength } \\Z=\text { atomic num. }\end{array}$$
$$\begin{array}{l}\text { * For a particular member of spectral line }\\\text { we have - } \\\qquad \begin{aligned}\frac{1}{\lambda} &\alpha z^{2} \quad \Rightarrow\left[\lambda \alpha \frac{1}{z^{2}}\right] \\& \Rightarrow \lambda=\frac{k}{z^{2}} \text { (let) }-(1) \\\Rightarrow & \lambda_{H}=k\end{aligned}\end{array}$$
$$\begin{array}{l}\text { For any other ionic atom for which } \\\text { wavelength }=\frac{\lambda_{H}}{4}=\frac{k}{4}-(2) \\\text { from eq (1) and (2) } \\\text { we get } \frac{k}{z^{2}}=\frac{k}{4}\\\Rightarrow z^{2}=4 \\\Rightarrow z=2\quad\left(\mathrm{He}^{+}\right) \text {atom. } \\\text { Hence, option (A) is correct. }\end{array}$$
The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of $$108.5nm$$. the ground state energy of an electron of this ion will be
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$$3.4eV$$
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$$13.6eV$$
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$$54.4eV$$
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$$122.4eV$$
Explanation
For third of Balmer series $$n_1=2,\ n_2=5$$
$$\therefore \dfrac{1}{\lambda}=RZ^2 \left[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2} \right]$$
gives $$Z^2=\dfrac{n_1^2 n_2^2}{(n_2^2-n_1^2) \lambda R}$$
On putting values $$Z=2$$
From $$E=-\dfrac{13.6Z^2}{n^2}=\dfrac{-13.6(2)^2}{(1)^2}=-54.4\ eV$$
If the kinetic energy of an electron in the first orbit of $$H$$ atom is $$13.6 eV$$. Then the total energy of an electron in the second orbit of $${He}^{+}$$ is:
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$$13.6 eV$$
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$$3.4 eV$$
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$$-13.6 eV$$
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$$-3.4 eV$$
Explanation
$$\begin{array}{l}\text { Given- * kinetic energy (KE) of e" in first orbit of } \\\text { H-atom is } 13.6 \mathrm{eV.} \\\text { we know that - } \\\qquad k E=13.6 \mathrm{z}^{2}/n^2 \quad n \rightarrow n^{\text {th }} \text { orbit } \\\quad z \rightarrow \text { atomic number. }\end{array}$$
$$\begin{array}{l}\text { we also know that - } \\\text { (Total Energy) }=-(\text { kinetic Energy) } \\\Rightarrow \quad T E=-K E\\\Rightarrow T E=-13.6 \frac{z^{2}}{n^{2}} e V \\\text { Now, for } H e^{+} \text {atom }(z=2) \text { in second } \text { orbit }(n=2)\\\qquad T E=-13.6 \text { eV } \frac{4}{4}=-13.6\mathrm{eV}\end{array}$$
Which of the following weighs the least?
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2 g atom of N (atomic wt of N =14)
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$$ 3 \times 10^{23} $$ atoms of C (atomic wt. of C=12)
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1 mole of S ( atomic weight of S = 32 )
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7 g silver ( atomic wt. of Ag =108)
In a hypothetical system a particle of mass $$m$$ and charge $$-3q$$ is moving around a very heavy particle having charge $$q$$. Assuming Bohr's model to be true to this system, the orbital velocity of mass $$m$$ when it is nearest to the heavy particle is
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$$\cfrac{3{q}^{2}}{2{\epsilon}_{0}h}$$
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$$\cfrac{3{q}^{2}}{4{\epsilon}_{0}h}$$
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$$\cfrac{3{q}^{}}{2{\epsilon}_{0}h}$$
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$$\cfrac{3{q}^{}}{4{\epsilon}_{0}h}$$
A positronium consists of an electron and a positron revolving about their common centre of mass. Calculate the separation between the electron and positron in their first excited state:
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$$0.529\ A$$
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$$1.058\ A$$
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$$2.116\ A$$
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$$4.232\ A$$
Explanation
$$\begin{array}{l}\text { Given, positronium consist of electron and positron }\end{array}$$
$$\begin{array}{l}\text { Let w be angular velocity about } 0 . \\\text { balancing force } \\\qquad \frac{k e^{2}}{4 R^{2}}=m \omega^{2} R\text { --- (1) }\end{array}$$
$$\begin{array}{l}\text { Now, Angular momentum } \\\qquad \begin{aligned}m(\omega R) R&=\frac{n h}{2 \pi} \\m \omega R^{2} &=\frac{n h}{2 \pi } ---(2)\end{aligned}\end{array}$$
$$\begin{array}{l}\text { from }(1) \text { and }(2) \\\quad 2 R=\frac{n^{2} h^{2}}{\pi^{2} m K e^{2}}\end{array}$$
$$\begin{array}{l}2 R=4 \times\left(\frac{n^{2} h^{2}}{4 \pi^{2} m K e^{2}}\right) \\2 R=4\times\frac{0.529 n^{2}}{z}\end{array}$$
$$\begin{aligned}&=4 \times 0.529 \dot {A} \quad, \quad n=1,z=1 \\2 R &=4.232 \dot{A}\end{aligned}$$
The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is :
(in $$A^o$$)
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1215
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1640
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2430
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4687
Calculate number of electrons present in $$500\ cm^{3}$$ volume of water.
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$$1.67\times 10^{20}$$
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$$1.67\times 10^{26}$$
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$$1.67\times 10^{23}$$
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$$3.2\times 10^{26}$$
Explanation
$$\begin{array}{l}\text { Given } \\\text { of electron in one molecule of }\mathrm{H}_{2}O=2+8 \\=10\end{array}$$
$$\begin{array}{l}\text { we know } \\\text { density of watere } 1 \mathrm{~g} /\mathrm{cm}^{3} \\\text { given volume }=500\mathrm{~cm}^{3} \\\text { No. of mole of water }=\left(\frac{500}{18}\right)\end{array}$$
$$\begin{array}{l}\text { Since } 1 \text { mole contain } 6.02 \times 10^{23} \text { molecule of } H_{2}O \\\text { and } 1 \mathrm { H}_{2}\mathrm{O} \text { contain } 10\mathrm{e}^{-}\end{array}$$
$$\begin{aligned}\qquad \begin{aligned}\text { total } e^{-} \text { contain } &=\frac{500}{18}\times 6.022 \times 10^{23}\times 10 \\&=1.67 \times 10^{26}\end{aligned}\end{aligned}$$
The magnitude of angular momentum, orbit radius and frequency of revolution of electron in hydrogen atom corresponding to quantum number $$n$$ are $$L,r$$ and $$f$$ respectivey. Then according to Bohr's theory of hydrogen atom
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$$f{r}^{2}L$$ is constant for all orbits
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$$f{r}^{}L$$ is constant for all orbits
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$${f}^{2}rL$$ is constant for all orbits
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$$fr{L}^{2}$$ is constant for all orbits
The electron in a hydrogen atom makes a transition $${n}_{1}\rightarrow {n}_{2}$$ whose $${n}_{1}$$ and $${n}_{2}$$ are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The frequency of orbital motion of the electron in the initial state is $$1/27$$ of that in the final state. The possible values of $${n}_{1}$$ and $${n}_{2}$$ are
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$${n}_{1}=4,{n}_{2}=2$$
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$${n}_{1}=3,{n}_{2}=1$$
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$${n}_{1}=8,{n}_{2}=1$$
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$${n}_{1}=6,{n}_{2}=3$$
The wavelength of the first member of the Balmer series in hydrogen spectrum is $$x$$ $$\mathring{A}$$. Then the wavelength (in $$\mathring{A}$$) of the first member of Lyman series in the same spectrum is
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$$\cfrac{5}{27}x$$
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$$\cfrac{4}{3}x$$
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$$\cfrac{27}{5}x$$
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$$\cfrac{5}{36}x$$
Explanation
$$ \begin{array}{l} \text { Balmer series : transition take place from } \\ n=2 \text { to } n=3,4,5, \text { so on } \\ \text { for first member of balmer } \\ n_{1}=2 \text { to } n_{2}=3 \end{array} $$
$$ \lambda=x \dot {A} $$
$$ \begin{array}{l} \frac{1}{\lambda}=R z^{2}\left(\frac{1}{n_{1} 2}-\frac{1}{n_{2}^{2}}\right) \\ \frac{1}{x}=R z^{2}\left(\frac{1}{4}-\frac{1}{9}\right) \\ \frac{1}{x}=R z^{2}\left(\frac{5}{36}\right)-(1) \end{array} $$
$$ \begin{array}{l} \text { For first member of Lyman } \\ \qquad \begin{aligned} n_{1} &=3 \quad n_{2}=2 \\ \frac{1}{\lambda} &=R_{2}^{2}\left(\frac{1}{1}-\frac{1}{4}\right) \\ \frac{1}{\lambda} &=R z^{2}\left(\frac{3}{4}\right) \end{aligned} \end{array} $$
$$ \begin{array}{l} \text { divide (i) } \div \text { (ii) } \\ \qquad \begin{array}{l} \frac{ \lambda }{x}=\left(\frac{5}{36}\right)\left(\frac{4}{3}\right) \\ \lambda=\left(\frac{5}{27}\right) x \end{array} \end{array} $$
In a hypothetical system a particle of mass m and charge -3q is moving around a very heavy particle having charge q. Assuming Bohr's model to be true to this system, The orbital velocity of mass m when it is nearest to heavy particle is
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$$\dfrac{3q^2}{2 \varepsilon_0h}$$
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$$\dfrac{4q^2}{2 \varepsilon_0h}$$
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$$\dfrac{3q}{2 \varepsilon_0 h}$$
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$$\dfrac{3q}{4 \varepsilon_0 h}$$
Explanation
$$\begin{array}{l}\text { Given, } \\\text { a hypothetical system } \\\text { Bohr's model is true by } \\\text { assuming. }\end{array}$$
$$\text { } \begin{aligned}I) F_{\text {centripetal }} &=\text { culoumb force } \\&\frac{m v^{2}}{r}=\frac{1 \times q \times+3 q}{4 \pi \varepsilon_{0} \times r^{2}} \\& m v r=\frac{3 q^{2}}{4 \pi \varepsilon_{0} v}\cdots(1)\end{aligned}$$
$$\begin{array}{l}\text { II) Angular momentum }=\frac{n h}{2 \pi} \\m v r=\frac{n h}{2\pi}\text { (ii) } n=\text { no of orbit } \\h=\text { plank constant }\end{array}$$
$$\begin{array}{l}(i)=(i i) \\\frac{n h}{2\pi}=\frac{3 q^{2}}{4 \pi \varepsilon_{0} v}\end{array}$$
$$\begin{array}{l}\text { for } n=1 \\\qquad\begin{array}{l}\frac{h}{2 \pi}=\frac{3 q^{2}}{4 \pi \varepsilon_{0} v} \\\text { velocity in } 1^{\text {st }} \text { orbit }(v)=\frac{3 q^{2}}{2 \varepsilon_{0} h}\end{array}\end{array}$$
An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its kinetic energy. Consequently the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The main KE of colliding electron will be
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$$10.2eV$$
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$$1.9eV$$
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$$12.1eV$$
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$$13.6eV$$
If Rutherford would have used $$\beta $$-particles instead of $$\alpha $$-particles in his Goldleaf experiment which of the following observation would definitely not have been made?
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Get condensed or concentrated
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Majority of $$\beta $$-particles passing unperfected through the foil
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Very few $$\alpha $$-particles getting absorbed
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Wave nature of electron
Explanation
Beta particles are heavier as compared to alpha particles. So if Rutherford's alpha particle experiment was done with beta particles the results would not have been the same. Not much deviation would have occurred and the proper theory related to the structure of the atom would have not been decided.
The ratio of areas within the electron orbits for the first excited state to the ground state for hydrogen atom is
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$$16:1$$
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$$18:1$$
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$$4:1$$
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$$2:1$$
Explanation
The square radius of the orbit is directly proportional to the power of four of the number of state.
$$r^2 \alpha$$ $$n^4$$
Where, n is the number of state
Areas of hydrogen for excited state
$$A_1$$ $$\alpha$$ $$n_1^4$$
Here, n = 2 for excited state
The area of hydrogen for the ground state
$$A_0$$ $$\alpha$$ $$n_0^4$$
Here, n = 1 for ground state
Now, the ratio of areas between the electron orbits for the first excited state to the ground state for the hydrogen atom is
$$\frac{A_1}{A_0} = (\frac{n_1}{n_0})^4$$
$$\frac{A_1}{A_0} = (\frac{2}{1})^4$$
$$\frac{A_1}{A_0} = (\frac{16}{1})$$
Hence, The ratio of the area is $$16:1$$.
The electric potential between a proton and an electron is given by $$V=VIn\dfrac { r }{ { r }_{ 0 } } $$, where $${ r }_{ 0 }$$ is a constant. Assuming Bohr's model to be applicable,write variation of $${ r }_{ n}$$ with $$n,\ n$$ being the principal quantum number.
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$${ r }_{ n }\propto n$$
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$${ r }_{ n }\propto \dfrac { 1 }{ n } $$
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$${ r }_{ n }\propto { n }^{ 2 }$$
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$${ r }_{ n }\propto \dfrac { 1 }{ { n }^{ 2 } } $$
Choose the correct statement (s) for hydrogen and deuterium atoms considering motion of reaction
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The radius of first Bohr orbit of deuterium is less than that of hydrogen
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The speed of electron in the first Bohr orbit of deuterium is more that that of hydrogen
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The wave length of first Balmer line of deuterium is more than that of hydrogen
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The angular momentum of electron in the first Bohr orbit of deuterium is more than that of hydrogen
Explanation
$$ \begin{array}{l} \text { Bohr's considerd the nucleus is at rest and only electron is revolving around it. } \\ \text { but here in question motion of nucleus is given.} \end{array} $$
$$ \begin{array}{l} \text { Hydrogen }= _{1}{H}^{2} \\ \text { deuterium }= _1{\mathrm{H}^{2}} \end{array} $$
$$ \begin{array}{l} \text { Now, if we consider motion of nucleus } \\ \text { also, then we have to consider nucleus mass } \\ \text { also. } \end{array} $$
$$ \text{ (a)} \text { Now, } r \propto \dfrac{n^{2}}{m} $$
$$ \begin{array}{l} \text { So, radius of deuterium is less than that of hydrogen, option (a) is correct } \end{array} $$
$$ \text { (b) velocity } \propto \dfrac{z^{2}}{n^{2} m} $$
$$ \begin{array}{l} \text { So, velocity of electron in hydrogen is more, option B incorrect } \end{array} $$
$$ \begin{array}{l} \text { (c) wavelength }\\ \dfrac{1}{\lambda}=R z^{2}\left(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}\right) \\ \text { and } \therefore R \propto m \end{array} $$
$$ \begin{aligned} & \frac{1}{\lambda} \propto R \\ &=\frac{1}{\lambda} \propto m \end{aligned} $$
$$ \begin{array}{l} \text { So, wavelength of hydrogen will be more than that of deuterium, } \end{array} $$
$$ \text { option c is incorrect } $$
$$ \begin{array}{l} \text { (d) as angular momentum is conserved always inspite of mass and atomic no. , it only depend on Energy level. } \\ \text { mvr }=\dfrac{n h}{2 \pi} \end{array} $$
$$ \text { So, option }(d) \text { is incorrect. } $$.
According to Bohr's theory of the hydrogen atom, the speed $$v_n$$ of the electron in a stationary orbit is related to the principal quantum number n as (C is a constant).
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$$v_n=C/n^2$$
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$$v_n=C/n$$
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$$v_n=C\times n$$
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$$v_n=C\times n^2$$
Assuming Bohr's model for $$ Li^{++} $$ atom, the first excitatio energy of ground state of $$ Li^{++} $$ atom is
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10.2 eV
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91.8 eV
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13.6 eV
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3.4 eV
Explanation
For $$Li^{++}$$ model,
$$2=3$$
In first excitation of ground state,
$$n_1=1$$ and $$n_2=2$$
In Bohr's model,
$$\Delta E=(13.6ev)2^2\left[\dfrac{1}{n_1^2}-\dfrac{1}{n^2_2}\right]$$
Here, $$\Delta E=(13.6)(3)^2\left[\dfrac{1}{1}-\dfrac{1}{4}\right]$$
$$\Delta E=(13.6)9\times \dfrac{3}{4}=91.8$$eV
So, first excitation energy for ground state in $$Li^{++}$$ is $$91.8$$eV
Option B is correct.
The energy levels of a certain atom for 1st, 2nd and 3rd levels are E, 4E$$_{1\to3}$$ and 2E respectively. A photon of wavelength $$\lambda $$ is emitted for a transition $$3\rightarrow 1$$. What will be the wavelength of emissions for transition $$2\rightarrow 1$$:
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$$\lambda /3$$
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$$4\lambda /3$$
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$$3\lambda /4$$
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$$3\lambda$$
Explanation
Let $$E_{1}, E_{2}, E_{3}$$ represents the energy of $$1st, 2nd$$ and $$3rd$$ energy levels respectively such that
$$E_{1}=E$$
$$E_{2}= 4E$$
$$E_{2}= 2E$$
Now, we know that during transition of electron from higher energy level to lower energy level, energy in released in the form of photon having energy equal to the energy difference between two energy levels.
Ob transition from $$3 \rightarrow 1$$ energy level.
Energy of photon released $$= E_{3}- E_{1} $$--------$$(1)$$
As we know that energy of photon is given by $$=\dfrac{hc}{\lambda}$$
where $$\lambda $$ wavelength of photon
from $$(1)$$
$$\dfrac{hc}{\lambda} = E_{3}-E_{1}=E$$
$$\Rightarrow \lambda= \dfrac{hc}{E}$$-------$$(2)$$
On transition from $$2 \rightarrow 1$$ energy level.
Energy of photon released $$=E_{2}-E_{1}$$--------$$(2)$$
Let, wavelength of photon released be $$\lambda$$
$$\Rightarrow $$ Energy of photon released $$=\dfrac{hc}{\lambda}$$
From $$(3)$$
$$E_{2}-E_{1}= \dfrac{hc}{\lambda ,}$$
$$\Rightarrow 3E= \dfrac{hc}{\lambda ,}$$
$$\Rightarrow \lambda ' = \dfrac{hc}{3E} =\dfrac{1}{3} \left( \dfrac{hc}{E} \right)$$
$$\Rightarrow \lambda ' = \dfrac{1}{3} (\lambda) (from \ (2))$$
$$\Rightarrow \lambda ' = \dfrac{\lambda}{3}$$
Hence, wavelength of photon emitted for transition $$2 \rightarrow 1$$ in $$\dfrac{\lambda}{3}$$.
Assuming Bohr's model for $$Li^ {++}$$ atom, the first excitation energy of ground state of $$Li^ {++}$$ atom is
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$$10.2 eV$$
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$$91.8 eV$$
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$$13.6 eV$$
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$$3.4 eV$$
Explanation
$$ \begin{array}{l} \text { Given, } \\ Li^{++} \text { atom: } \\ \text { first excitation Energy of ground state } L i^{++} \\ \text {mean excitation of electrom from } \\ \qquad n=1 \text { to } n=2 \end{array} $$
$$ \begin{array}{l} \text { we know, } \\ \text { Energy }=13.6 z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \end{array} $$
$$ \begin{aligned} &=13.6 \times 9 \times\left(\frac{1}{1}-\frac{1}{4}\right) \\ E &=\frac{13.6 . \times 9 \times 3}{4} \mathrm{ev} \\ E &=91.8 \mathrm{ev} \end{aligned} $$
In a hypothetical system a particle of mass m and charge -3 q is moving around a very heavy particle having charge q. Assuming Bohr's model to be true to this system, the orbital velocity of mass m when it is nearest to heavy particle is:
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$$\frac{3q^2}{2\varepsilon _0h}$$
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$$\frac{3q^2}{4\varepsilon _0h}$$
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$$\frac{3q}{2\varepsilon _0h}$$
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$$\frac{3q}{4\varepsilon _0h}$$
A hydrogen atoms emits green light when it changes from the n=4 energy level to n=2 level. which color of light would the atom emit when it changes from the n=5 level to n=2 level ?
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red
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yellow
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green
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violet
Explanation
According to Balmer series.
$$\overline { V } =R\left[ \dfrac { 1 }{ { z }^{ 2 } } -\dfrac { 1 }{ { n }^{ 2 } } \right] $$
$$n=3,4,5,6,......$$
Spectra basically is belonging the location of visible zone. When wavelength is in between $$4000{ A }^{ 0 }$$ to $$5000{ A }^{ 0 }$$ then, green color of light would the atom emit. When it changes from the $$n=5$$ level to $$n=2$$ level.
If the ionisation energy of a hydrogen like Bohr atom is 4 Rydberg, then the wavelengths of radiation emitted when the electron jumps from first excited state to the ground state and the radius of the first orbit of this atom are:
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$$3.04 \overset { \circ }{ A } ,0.53 \overset { \circ }{ A } $$
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$$304 \overset { \circ }{ A } , 0.265 \overset { \circ }{ A } $$
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$$912 \overset { \circ }{ A } , 0.53 \overset { \circ }{ A } $$
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$$3.04 \overset { \circ }{ A } , 0.265 \overset { \circ }{ A } $$
Explanation
$$\begin{array}{l}\text { Ionisation energy, } E=R z^{2}\\\qquad\begin{aligned}Rz^{2} &=4 R \\z &=2\end{aligned}\end{array}$$
$$\text { Radius of 1st orbit is } \dfrac{0.529}{2}A^{0}\text{=0.265} A^{0}$$
$$\begin{array}{l}\text { wavelength } \\\qquad \dfrac{1}{\lambda}=R z^{2}\left[\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}\right]\end{array}$$
$$\begin{array}{ll}n_{1}=1 & n_{2}=2 \\\text { we get, } & \lambda=304 A^{0}\end{array}$$
Find an expression for de Broglie wavelength for an electron in the $$n^{th}$$ orbit of the Bohr model of the hydrogen atom and hence find its value when the electron is in n = 4 level.
Take radius of nth orbital of Bohr's model of H-atom; $$r_n = (53 \ n^2) $$ pm
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1.33 nm
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1.20 nm
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1.50 nm
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None
The innermost orbit of the hydrogen atom has a diameter of $$1.06 A^o$$. What is the diameter of the tenth orbit?
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$$5.3 A^o$$
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$$10.6 A^o$$
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$$53 A^o$$
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$$106 A^o$$
An electron jumps from the $$4^{th}$$ orbit to the $$2^{nd}$$ orbit of hydrogen atom $$(R=10^{5}\ Cm^{-1})$$. Frequency is in $$Hz$$ of emitted radiation will be
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$$\dfrac {3\ \times 10^{5}}{16}$$
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$$\dfrac {3\ \times 10^{15}}{16}$$
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$$\dfrac {9\ \times 10^{15}}{16}$$
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$$\dfrac {9\ \times 10^{5}}{16}$$
Explanation
$$\begin{array}{l}\text { Electron jumps from } 4^{\text {th }} \text { orbit to } \\2^{\text {nd }} \text { orbit. } \\\text { Let wavelength be "x" then } \\\qquad \frac{1}{\lambda}=R Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\end{array}$$
$$\text { Here, } \begin{aligned}& R=10^{5} \mathrm{~cm}^{-1} \\& z=1\\& n_{1}=2, n_{2}=4 \\\frac{1}{\lambda} &=10^{5}\left[\frac{1}{2^{2}}\frac{1}{4^{2}}\right] \\\frac{1}{\lambda}&=10^{5}\left[\frac{1}{4}-\frac{1}{16}\right] \\\frac{1}{\lambda} &=10^{5}\left[\frac{3}{16}\right] \\\frac{1}{\lambda} &=\frac{3 \times 10^{5}}{16}-(1)\end{aligned}$$
$$\begin{array}{l}\text { Frequency is } \frac{c}{\lambda} \\c=3 \times 10^{8} \mathrm{~m}=3 \times 10^{10} \mathrm{~cm} \\\text { From frequency, } v=3 \times 10^{10} \times \dfrac{3 \times 10^{5}}{16}\\\qquad v=\dfrac{9 \times 10^{15}}{16}\end{array}$$
In a Geiger-Marsden experiment if the distance of closest approach is $$d$$ to the nucleus and an $$\alpha$$ particle of energy E then
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$$d \propto E$$
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$$d \propto \frac{1}{E}$$
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$$d^2 \propto E$$
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$$d^2 \propto \frac{1}{E}$$
The $$H_{a}$$-line of hydrogen
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has a wavelength 4860 $$\overset {0}{A}$$
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has a wavelength 6562$$\overset {0}{A}$$
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has a wavelength smaller then that of the $$H_{\beta}$$-line
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is emitted in the transition from the second excited state of the first excited state
According to Bohr modal, magnetic field at the centre (at the nucleus) of a hydrogen atom due to the motion of electron in $$n^{th}$$ orbit is proportional to:
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$$1/n^3$$
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$$n^3$$
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$$n^5$$
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1/$$n^5$$
The wave number of first line in Balmer series of hydrogen spectrum is (Rydberg constant, $$R_H = 109,687 \,cm^{-1}$$) nearly
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$$82,259 \,cm^{-1}$$
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$$152,00 \,cm^{-1}$$
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$$97,492 \,cm^{-1}$$
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$$15,200 \,cm^{-1}$$
Explanation
$$\begin{array}{l}\text { Wave number } \\\text { Here, }\quad\left[\bar{v}=R_{H} z^{2}\left[\frac{1}{n_{i}{ }^{2}}-\frac{1}{n_{2}^{2}}\right]\right. \\R_{H}=109687 \\z=1\end{array}$$
$$ \begin{aligned}\text { Balmar } n_{1}=2 \text { , and } n_{2}=3\\\bar{v}=109687\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right]\\\bar{v}=109687\left[\frac{1}{4}-\frac{1}{9}\right]\end{aligned}$$
$$\bar{v}=15200 \mathrm{~cm}^{-1}$$
Which of the following is inversely proportional to the square of principal quantum number in Bohr's atomic model?
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Total energy of the electron in $$n^{th}$$ orbit.
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Radius of the $$n^{th}$$ orbit.
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Frequency of revolution in $$n^{th}$$ orbit.
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Time period of revolution in $$n^{th}$$ orbit.
Choose the correct statements for hydrogen and deuterium atoms ( considering motion of nuclie as )
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The radius of first Bohr orbit of deuterium is less than that of hydrogen
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The speed of electron in the first Bohr orbit of deuterium is more than that of hydrogen.
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The wavelength of first Balmer line of deuterium is more than that of hydrogen
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The angular momentum of electron in the first Bohr orbit of deuterium is more than that of hydrogen.
The ratio of the radii of the electron orbits for the first excited state to the ground state for the hydrogen atom is:
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2 : 1
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4 : 1
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8 : 1
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16 : 1
The wave number of first line of Balmer series of hydrogen atom is $$15200\ cm^{-1}$$. What is the wave number of first line of Balmer series of $$Li^{2+}$$ ion.
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$$15200\ cm^{-1}$$
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$$136800\ m^{-1}$$
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$$76000\ cm^{-1}$$
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$$13680\ cm^{-1}$$
Explanation
As we know the energy of constant state for hydrogen species like $$Li^+2$$ or $$He^+$$ are multiple of integral type of $$Z^2$$ times the energy of hydrogen atom in stationary form$$,$$
Hence the wave number of first line of ion $$= 3^2 \times 15200 / cm$$
$$= 9 \times 15200 / cm = 136800 / cm$$
hence,
option $$(B)$$ is correct answer.
In a Bohr atom the electron is replaced by a particle of mass 150 times the mass of the electron an the same charge, If $$a_0$$ is the radius of the first Bohr orbit of the orbital atom, then that of the new atom will be
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$$150 a_0$$
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$$\sqrt {150 a_0}$$
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$$\sqrt {\dfrac {a_0}{150}}$$
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$$\dfrac {a_0} {150}$$
The ionization potential of H-atom is $$13.6$$V. The H-atoms in ground state are excited by monochromic radiations of photon energy $$12.09$$ eV. Then the number of spectral lines emitted by the excited atoms will be:
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$$1$$
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$$2$$
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$$3$$
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$$4$$
A $$12.5 eV$$ electron beam is used to bombard gaseous hydrogen at room temperature. What series of spectral line will be emitted ?
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$$1 $$ Lyman, $$2$$ Balmer
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$$2$$ Lyman, $$1$$ Balmer
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$$1$$ Lyman, $$1$$ Balmer, $$1$$ Paschen
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$$3$$ Balmer
The wavelength of the first member of the Balmer series is $$656.3\ nm$$. The wavelength of the second line of the Lyman series is
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$$81.02\ nm$$
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$$121.6\ nm$$
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$$364.8\ nm$$
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$$729.6\ nm$$
The speed of sound in air is $$v$$. Both the source and observer are moving towards each other with equal speed $$u$$. The speed of wind is$$w$$ from source to observer. Then, the ratio $$(\cfrac{f}{{f}_{0}})$$ of the apparent frequency to the actual frequency is given by
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$$\cfrac{v+u}{v-u}$$
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$$\cfrac{v+w+u}{v+w-u}$$
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$$\cfrac{v+w+u}{v-w-u}$$
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$$\cfrac{v-w+u}{v-w-u}$$
Explanation
$$\begin{array}{l} f=\left( { \dfrac { { v+u+w } }{ { v-u+w } } } \right) { f_{ 0 } } \\ \Rightarrow \dfrac { f }{ { { f_{ 0 } } } } =\left( { \dfrac { { v+u+w } }{ { v-u+w } } } \right) \\ Hence, \\ option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$$
According to Moseley's law, the ratio of the slope of graph between $$\sqrt { f } $$ and $$Z$$ for $${ k }_{ \beta }$$ and $${ k }_{ \alpha }$$ is
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$$\sqrt { \dfrac { 32 }{ 27 } } $$
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$$\sqrt { \dfrac { 27 }{ 32 } } $$
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$$\sqrt { \dfrac { 5 }{ 56 } } $$
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$$\sqrt { \dfrac { 36 }{ 5 } } $$
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Practice Class 12 Medical Physics Quiz Questions and Answers
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