A stream of uncharged particles

MCQ Questions for Class 12 Medical Physics Atoms Quiz 8

Linear momentum of an electron in Bohr orbit of H-atom (principal quantum number

$n$ ) is proportional to :

0%
$\dfrac { 1 }{ { n }^{ 2 } }$
0%
$\dfrac { 1 }{ n }$
0%
$n$
0%
${ n }^{ 2 }$

Explanation

The angular momentum in a Bohr orbit is given as

$L=mvr=\dfrac{nh}{2\pi}$

$\implies mv=\dfrac{nh}{2\pi r}$

$\implies p=mv\propto n$

The nuclear radius of

$_{ 4 }^{ }{ { Be }^{ 8 } }$ nucleus is :

0%
$1.3\times { 10 }^{ -15 }m$
0%
$2.6\times { 10 }^{ -15 }m$
0%
$1.3\times { 10 }^{ -13 }m$
0%
$2.6\times { 10 }^{ -13 }m$

Explanation

Mass number of

$_4Be^8$

$A = 8$

Nuclear radius,

$R = R_o A^{1/3}$ where

$R_o = 1.3 \times 10^{-15} m$

$\therefore$

$R = 1.3\times 10^{-15} \times 8^{1/3}$

$m$

$\implies$

$R = 1.3\times 10^{-15} \times 2 = 2.6\times 10^{-15} m$

The characteristic spectrum of an atom is observed as

0%
Pure line spectrum
0%
Emission band spectrum
0%
Absorption line spectrum
0%
Absorption band spectrum

The cathode rays are

0%
A stream of electrons
0%
A stream of positive ions
0%
A stream of uncharged particles
0%
The same as canal rays

Atomic spectrum should be :

0%
Pure line spectrum
0%
Emission band spectrum
0%
Absorption line spectrum
0%
Absorption band spectrum

The ratio of the radii of the first three Bohr Orbits is

0%
$1:\displaystyle \frac{1}{2}:\frac{1}{3}$
0%
$1:2:3$
0%
$1:4:9$
0%
$1:8:27$

Explanation

Radius of

$n^{th}$ Bohr orbit,

$r_n = 0.529 \dfrac{n^2}{Z}$

$A^o$

$\implies$

$r_n \propto n^2$

Thus ratio of first three Bohr orbit is

$r_1 : r_2 : r_ 3 = 1^2 : 2^2 : 3^ 2$

$\implies$

$r_1 : r_2 : r_ 3 = 1 : 4 : 9$

In Hydrogen atom, which of the following transitions produces a spectral line of maximum frequency.

0%
$2\rightarrow 1$
0%
$6\rightarrow 2$
0%
$4\rightarrow 3$
0%
$5\rightarrow 2$

Explanation

The Rydberg formula for Hydrogen atom,

$\dfrac{1}{\lambda}=R[1/n_1^2-1/n_2^2]$ where

$R=$ Rydberg constant and

$n_1<n_2$

For

$2\rightarrow 1$ ,

$\dfrac{1}{\lambda_{21}}=R[1/1^2-1/2^2]=(3/4)R=0.75 R$ ;

For

$6\rightarrow 2$ ,

$\dfrac{1}{\lambda_{62}}=R[1/2^2-1/6^2]=(2/9)R=0.22 R$ ;

For

$4\rightarrow 3$ ,

$\dfrac{1}{\lambda_{43}}=R[1/3^2-1/4^2]=0.05R$ and

For

$5\rightarrow 2$ ,

$\dfrac{1}{\lambda_{52}}=R[1/2^2-1/5^2]=0.21R$

Since frequency is inversely proportional to wavelength so

$2\rightarrow 1$ transition will give the maximum frequency.

As the electron in Bohr's orbit of hydrogen atom passes from state

$n=2$ to,

$n=1$ , the kinetic energy

$(K)$ and the potential energy

$(U)$ changes as

0%
K four fold, U also four fold
0%
K two fold, V also two fold
0%
K four fold, U two fold
0%
K two fold, U four fold

Explanation

Given,

Electron in Bohr's orbit of hydrogen atom passes from state

$n=2$ to

$n=1$ ,

KE of an electron in nth orbit :

$K_n \propto \dfrac{1}{n^2}$ and PE of an electron in nth orbit :

$U_n\propto \dfrac{1}{n^2}$

$\therefore$ When an electron passes from state n = 2 to n = 1

$\dfrac{K_2}{K_1}=\dfrac{1^2}{2^2}=\dfrac{1}{4}$
or

$K_1=4K_2$

$\dfrac{U_2}{U_1}=\dfrac{1^2}{2^2}=\dfrac{1}{4}$
or

$U_1=4U_2$

The number of de Broglie waves of an electron in the

${ n }^{ th }$ orbit of an atom is :

0%
$n$
0%
$n-1$
0%
$n+1$
0%
$2n$

Explanation

The only places an electron can take in an atom are those where it can complete a multiple of a full wavelength. So when an electron (in wave form) starts from a certain point it must come back to the same point and be at the exact same position as when it started from that point. Otherwise, the wave wouldn't coincide with itself, and it would consequently interfere with itself and possibly destroy itself. This can't happen according to the laws of the conservation of matter and energy, so the electron only takes the orbits in which it can complete a number of full wavelengths. The number of waves is thus

$n$ .

A singly ionized helium atom in an excited state

$\left( n=4 \right)$ emits a photon of energy

$2.6 eV$ . Given that the ground state energy of hydrogen atom is

$-13.6 eV$ , the energy

$\left( { E }_{ f } \right)$ and quantum number

$\left( n \right)$ of the resulting state are respectively,

0%
${ E }_{ f }=-13.6eV$ , $n=1$
0%
${ E }_{ f }=-6.0eV$ , $n=3$
0%
${ E }_{ f }=-6.0eV$ , $n=2$
0%
${ E }_{ f }=-13.6 eV$ , $n=2$

Explanation

From the Bohr's theory of single electron species, we have the Energy of the state (quantum number = n) and atomic number = Z as

$E=-13.6\frac{Z^2}{n^2}$

Thus, Energy of

$4^{th}$ state in Helium ion is

$E_4^{He^+} = -13.6\times \frac{2^2}{4^2}=-3.4eV$

Energy of the emitted photon is

$E_p = 2.6eV$

Energy after emitting the photon is

$E_f = E_4^{He^+}-E_p = -3.4-2.6=-6eV$

The quantum number is given by

$n = (\frac{-13.6\times Z^2}{E_n})^{\frac{1}{2}} = \sqrt{\frac{-13.6\times 4}{-6}}=\sqrt{9}=3$

An electron in hydrogen atom after absorbing an energy photon jumps from energy state

${n}_{1}$ to

${n}_{2}$ . Then it returns to ground state after emitting six different wavelengths in emission spectrum. The energy of emitted photons is either equal to less than the absorbed photons. The

${n}_{1}$ and

${n}_{2}$ are

0%
${n}_{2}=4,{n}_{1}=3$
0%
${n}_{2}=5,{n}_{1}=3$
0%
${n}_{2}=4,{n}_{1}=1$
0%
${n}_{2}=5,{n}_{1}=1$

Explanation

From

$n_2=4$ ,

6 lines are obtained in emission spectrum.

Now,

$E_{4-2}=E_{absorbed}$

$F_{4-3}<E_{absorbed}$

and

$E_{4-1}, E_{3-1}, E_{2-1}>E_{absorbed}$

Hence,

$n_1=2$ and

$n_2=4$

The transition from from the state

$n = 5$ to

$n = 1$ in a hydrogen atom results in

$UV$ radiation. Infrared radiation will be obtained in the transition.

0%
$2\rightarrow 1$
0%
$3\rightarrow 2$
0%
$4\rightarrow 3$
0%
$6\rightarrow 2$

Explanation

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$

for UV

$n_1=1$ and

$n_2=5$ ,

$\lambda=\dfrac{1.04}{R}$

for infrared the wavelength must be greater the UV wavelength.

For

$n_1=3$ and

$n_2=4$ ,

$\lambda=\dfrac{7.2}{R}$ . This is the largest wavelength we can get

A gas of monoatomic hydrogen is bombarded with a stream of electrons that have been accelerated from rest through a potential difference of

$12.75V$ . In the emission spectrum one cannot observe any line of

0%
Lyman series
0%
Balmer series
0%
Paschen series
0%
Pfund series

Explanation

Energy of electron stream

$=12.75eV.$

The ground state electron having energy

$-13.6eV$ will acquire this energy and have total energy

$=13.6+12.75=-0.85eV$ .

This corresponds to the energy of electron in the 4th orbit. So, it will radiate energy when electron jumps from 4th orbit.

Hence, Pfund series cannot be observed.

In Bohr's theory of Hydrogen atom, the electron jumps from higher orbit 'n' to lower orbit 'p'. The wavelength will be minimum for the transition

0%
n = 5 to p = 4
0%
n = 4 to p = 3
0%
n = 3 to p = 2
0%
n = 2 to p = 1

Explanation

Wavelength of light emitted due to transition from

$n$ to

$p$ orbits

$\dfrac{1}{\lambda } = R \bigg(\dfrac{1}{p^2} - \dfrac{1}{n^2} \bigg)$

$\implies$

$\lambda = \dfrac{n^2 p^2}{R(n^2 - p^2)}$

Case (A) :

$n =5$

$p =4$

We get

$\lambda_A = \dfrac{5^2 \times 4^2}{R(5^2 - 4^2)} = 44.44 R$

Case (B) :

$n =4$

$p =3$

We get

$\lambda_B = \dfrac{4^2 \times 3^2}{R(4^2 - 3^2)} = 20.57 R$

Case (C) :

$n =3$

$p =2$

We get

$\lambda_C = \dfrac{3^2 \times 2^2}{R(3^2 - 2^2)} = 7.2 R$

Case (D) :

$n =2$

$p =1$

We get

$\lambda_D = \dfrac{2^2 \times 1^2}{R(2^2 - 1^2)} = 1.33 R$

Thus minimum wavelength is emitted when a transition takes place from

$n=2$ to

$p=1$ orbit.

An electron enters a parallel plate capacitor with horizontal speed

$\mu$ and is found to deflect by angle

$\theta$ on leaving the capacitor as shown. It is found that

$\tan\theta=0.4$ and gravity is negligible. If the initial horizontal speed is doubled, then

$\tan$ will be.

0%
$0.1$
0%
$0.2$
0%
$0.8$
0%
$1.6$

Explanation

let the horizontal length of plate be L

now

$t=L/\mu$

now

$v_y=at=\dfrac{eE}{m}\dfrac{L}{\mu}$

$tan\theta=\dfrac{v_y}{\mu}=\dfrac{eEL}{m\mu^2}=0.4$

when speed becomes double

$tan\theta=\dfrac{v_y}{\mu}=\dfrac{eEL}{m(2\mu^2)}=0.1$

The ionization energy of

${ Li }^{ 2+ }$ is equal to

0%
$9 hcR$
0%
$6 hcR$
0%
$2 hcR$
0%
$hcR$

Explanation

Ionization energy of an atom

$=hc R{ Z }^{ 2 }$
where

$R$ is the Rydberg constant and

$Z$ is the atomic number.
We know,

$Z=3$ for

${ Li }^{ 2+ }$
Ionization energy of

$Li^{2+}$

$={ \left( 3 \right) }^{ 2 }hcR$

$=9 hcR$

Given mass number of gold = 197,
Density of gold = 19.7 g/cm

$^3$
Avogadro's number =

$6 \times 10^{23}$ . The radius of the gold atom is approximately

0%
$1.7 \times 10^{-9} m$
0%
$1.5 \times 10^{-8} m$
0%
$1.5 \times 10^{-10} m$
0%
$1.5 \times 10^{-12} m$

Explanation

Volume occupied by one gram atom of gold

$=\dfrac{m}{\rho}= \dfrac{197 g}{19.7 g/cm^3} = 10 cm^3$

Volume of one atom

$=\dfrac{10}{N_A}= \dfrac{10}{6 \times 10^{-23}}$

$= \dfrac{5}{3} \times 10^{23} cm^3$
Let

$r$ be the radius of the atom.

$\therefore \ \dfrac{4}{3} \pi r^3 = \dfrac{5}{3} \times 10^{23}$

$r^3 = \dfrac{50 \times 10^{-24}}{4 \times 3.14}$

$r = 1.5 \times 10^{-10} m$

The wavelengths of the lines emitted in the Lyman series of the spectrum of hydrogen atom correspond to transitions between energy levels with total quantum numbers :

0%
$n = 3$ to $n = 1$
0%
$n = 3$ to $n = 2$
0%
$n = 4$ to $n = 1$
0%
$n = 4$ to $n = 2$

Explanation

Wavelength of the lines emitted lie in the Lyman series if the transition of electron takes place from higher energy levels

$n>1$ to ground state energy level

$n=1$ .
So, when transition of electron take place either from

$n=3$ to

$n=1$ or

$n=4$ to

$n=1$ , then wavelengths of lines so emitted lie in Lyman series.
Whereas transitions either from

$n=3$ to

$n=2$ or

$n=4$ to

$n=2$ correspond to Balmer series.

Rutherford's experiment on scattering of

$\alpha$ -particles showed for the first time that the atom has:

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Electrons
0%
Protons
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Neutrons
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Nucleus

The ratio of the wavelength for

$2\rightarrow 1$ transition in

${ Li }^{ + },{ He }^{ + }$ and

$H$ is :

0%
$1:2:3$
0%
$1:4:9$
0%
$4:9:36$
0%
$3:2:1$

Explanation

Using

$\cfrac { 1 }{ \lambda } =R{ Z }^{ 2 }\left( \cfrac { 1 }{ { n }_{ 1 }^{ 2 } } -\cfrac { 1 }{ { n }_{ 2 }^{ 2 } } \right) \Rightarrow \lambda \propto \cfrac { 1 }{ { Z }^{ 2 } }$

$\Rightarrow { \lambda }_{ Li }:{ \lambda }_{ He }:{ \lambda }_{ H }=\cfrac { 1 }{ 9 } :\cfrac { 1 }{ 4 } :\cfrac { 1 }{ 1 } =4:9:36$

Light of wavelength

$\lambda = 4000 A^o$ and intensity

$100 W/m^2$ is incident on a plate the threshold frequency

$5.5 \times 10^{14} Hz.$ Find the number of photons incident

$m^2$ per sec.

0%
$10^{21}$
0%
$3.0 \times 10^{19}$
0%
$2.02 \times 10^{20}$
0%
$2.02 \times 10^{21}$

Explanation

Let number of photons emitted per second be

$n.$

Then intensity

$= 100 W/m^2$

$\dfrac {nhc}{\lambda} = 100$

$n = \dfrac {100 \times \lambda }{hc}$

$= \dfrac {100 \times 4000 \times 10^{-10}}{6.6 \times 10^{-34} \times 3 \times 10^8 }$

$= 2.02 \times 10^{20}$

Which of the following is true?

0%
Lyman series is a continuous spectrum
0%
Paschen series is a line spectrum in the infrared
0%
Balmer series is a line spectrum in the ultraviolet
0%
The spectral series formula can be derived from the Rutherford model of the hydrogen atom

If the atom

$^{257}_{100}{Fm}$ follows the Bohr model and the radius of

$^{257}_{100}{Fm}$ is n times the Bohr radius, then the value of n is :

0%
4
0%
$\dfrac{1}{4}$
0%
200
0%
100

Explanation

According to Bohr's model

$r_m=\dfrac{m^2}{2}.r_0$

where, m is orbit number and

$r_0$ is the Bohr's radius for

$^{257}_{100}{FM}$ ,

$Z=100, m=5, r_m=nr_0$

$nr_0=\dfrac{5^2r_0}{100}$ or

$n=\dfrac{1}{4}$

The nucleus of an atom was discovered due to the experiment carried out by:

0%
Bohr
0%
Rutherford
0%
Moseley
0%
Thomson

If the series limit of Lymen series for hydrogen atom is equal to the series limit of Balmer series for a hydrogen like atom, then atomic number of this hydrogen like atom is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

By using,

$\dfrac {1}{\lambda} = RZ^{2} \left [\dfrac {1}{n_{1}^{2}} - \dfrac {1}{n_{2}^{2}}\right ]$
For hydrogen atom

$\dfrac {1}{(\lambda_{min})_{H}} = R\left [\dfrac {1}{1^{2}} - \dfrac {1}{\alpha}\right ] = R$

$\dfrac {1}{(\lambda_{min})_{H}} = R$ or

$(\lambda_{min})_{H} = \dfrac {1}{R} .... (i)$
For another atom

$\dfrac {1}{(\lambda_{min})_{atom}} = RZ^{2} \left (\dfrac {1}{2^{2}} - \dfrac {1}{\alpha}\right ) = \dfrac {RZ^{2}}{4}$

$\dfrac {1}{(\lambda_{min})_{atom}} = \dfrac {4}{RZ^{2}} .... (ii)$
From Eqs. (i) and (ii), we get

$\dfrac {1}{R} = \dfrac {4}{RZ^{2}} = Z = 2$ .

The adjoining diagram shows the energy levels for an electron in a certain, which transition shown represents the emission of a photon with the maximum energy

0%
I
0%
II
0%
III
0%
IV

Explanation

Energy released during emission of photon is maximum for transition III.
Energy of photon,

$E={ E }_{ 2 }-{ E }_{ 1 }=\left( -\cfrac { { E }_{ 1 } }{ { (2) }^{ 2 } } \right) -(-{ E }_{ 1 })=\cfrac { 3{ E }_{ 1 } }{ 4 }$

In an electron gun, the control grid is given negative potential, relative to cathode in order to.

0%
accelerate the electrons
0%
decrease KE of electrons
0%
repel the electrons
0%
decelerate the electrons

A radio transmitter operates at a frequency of 1000 kHz and a power of 66 kW. Then, the number of photons emitted per second is

0%
$1000$
0%
$10^{20}$
0%
$10^{12}$
0%
$10^{29}$

Explanation

$Number\ of\ photons\ emitted\ per\ second\ = \dfrac{Power}{Energy\ of\ photons}$

$= \dfrac{P}{h\nu} = \dfrac{66\times 1000}{6.6\times10^{-34}\times 10^{6}} = 10^{29}$

According to the Bohr's atomic model, the relation between principal quantum number (n) and radius of orbit(r) is?

0%
$r\propto n^2$
0%
$r\propto \displaystyle\frac{1}{n^2}$
0%
$r\propto \displaystyle \frac{1}{n}$
0%
$r\propto n$

Explanation

Electron angular momentum about the nucleus is an integer multiple of

$\displaystyle\frac{h}{2\pi}$ , where h is Planck's constant.
Thus,

$I\omega =mvr=\displaystyle \frac{nh}{2\pi}$
Hence

$r\propto n$ .

Which of the following expressions has the correct units to represent the radius of a hydrogen atom in its ground state?

0%
$\varepsilon_0h^2/\pi m_ee^2$
0%
$h^2m_ee^2/4\pi \varepsilon_0$
0%
$m_ec^2/h^2e^4$
0%
$e^4m_e/8\varepsilon^2_0h^2$

Explanation

Radius of nth orbit of hydrogen atom is given by

$\displaystyle r_n=\frac{n^2}{m_e}\left(\displaystyle\frac{h}{2\pi}\right)^2\left(\displaystyle\frac{4\pi \varepsilon_0}{e^2}\right)$ .
For ground state,

$n=1$

$\therefore \displaystyle r_1=\frac{\varepsilon_0h^2}{\pi m_ee^2}$ .

The ratio of kinetic energy and the total energy of the electron in the nth quantum state of Bohr's atomic model of hydrogen atom is

0%
$-2$
0%
$-1$
0%
$+2$
0%
$+1$

Explanation

The kinetic energy of the electron in the nth state

$K=\dfrac {mZ^2e^4}{8\varepsilon ^2_0h^2n^2}$
The total energy of the electron in the nth state

$T=\dfrac {-mZ^2e^4}{8\varepsilon ^2_0h^2n^2}$

$\therefore \dfrac {K}{T}=-1$

Energy of each orbit is

0%
changed
0%
fixed
0%
not same
0%
effected

The ratio of areas of the electron orbits for the first excited state and the ground state for the hydrogen atom is

0%
4 : 1
0%
16 : 1
0%
8 : 1
0%
2 : 1

Ratio of minimum to maximum wavelength in Ballmer series is :

0%
0.214583333
0%
0.233333333
0%
1:04
0%
3:04

Explanation

$\begin{array}{l}\text { For maximum, the ray should come from the } \\\text { just one level above the Balmer series i.e. } \\\text { Paschenn serios. } \\\text { for minimum the ray should come from } \\\text { infinity to Balmer series. }\end{array}$

$\begin{array}{l}\lambda=\text { wavelength } \\R=\text { Rydberg Constant } =1.01\times10^{7}\mathrm{~m}^{-1}\\n=\text { Respective}\text { series }\end{array}$

$\begin{array}{l}\text { } \\\text {for Lymen, n }=1\\\text {for Balmer, n }=2 \\\text {for Paschenn, n }=3 \\\text {for Brackett, n }=4 \\\text {for Pfund, n }=5\end{array}$

$\begin{array}{l}\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \\\text { for maximum } \\\qquad \begin{array}{l}\frac{1}{\lambda_{1}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) \\\text { For minimum } \\\frac{1}{\lambda_{2}}=R\left(\frac{1}{2^{2}}-\frac{1_{}^{2}}{\infty^2}\right)=R\frac{1}{2^{2}}\end{array}\end{array}$

$\begin{array}{l}\text { Ratio of Minimum to maximum = }\frac{\lambda_{2}}{\lambda_{1}}=0.214583333\end{array}$

With increasing quantum number, energy difference between adjacent levels in atoms :

0%
decreases
0%
increases
0%
remain constant
0%
zero

Explanation

$\begin{array}{l}\text { With increasing quantum}\text { number, energy difference } \\\text { between adjacent levels}\text { in atoms decreases. }\end{array}$

Velocity of electron in second Bohr's orbit as compared to velocity in first orbit is :

0%
equal
0%
one half
0%
2 times
0%
one fourth

Explanation

Velocity of electron in nth orbit is given by

$V_{n} = \frac{Ze^{2}}{2nh}$

$Z$ - atomic number

$e$ - charge on electron

$n$ - no. of orbit

$h$ - planks constant.

The velocity of electron in the 2nd Bhors orbit composed to first is one half.

According to Bohr's atomic model, angular momentum of electron in nth orbit is equal to an integral multiple of

0%
$2h/\pi$
0%
$h/2\pi$
0%
$h\pi/2$
0%
$h/\pi$

Explanation

$\text{ Angular momentum of an electron in the}$

$n^{\text {th }}\text{orbit is}$

$p=\frac{n h}{2 \pi}$

$h \rightarrow \text{planck's constant}$

$p=n \times \frac{h}{2 \pi}$

$n \rightarrow \text{ineger}$

$\therefore P\text{ is integral multiple of $\dfrac{h}{2 \pi}$}$

Bohr's atomic model is based upon :

0%
Einstein's relativistic theory
0%
classical theory
0%
planks quantum theory
0%
both b and c

In scattering experiment, find the distance of closest approach, if a

$6\ MeV\ \alpha - particle$ is used.

0%
$3.2\times 10^{-16}m$
0%
$2\times 10^{-14} m$
0%
$4.6\times 10^{-15} m$
0%
$3.2\times 10^{-15} m$

Explanation

Distance of closet approach of an

$\alpha$ -particle is given by,

$r=\dfrac { 2kZ{ e }^{ 2 } }{ { K }_{ \alpha } }$
ATQ,

$Z=79$

${ K }_{ \alpha }=6\quad MeV=6\times 1.6\times { 10 }^{ -13 }J$

Substituting these values we get,

$r=\dfrac { 2\times 9\times { 10 }^{ 9 }\times 79\times { \left( 1.6\times { 10 }^{ -19 } \right) }^{ 2 } }{ 6\times 1.6\times { 10 }^{ -13 } } \\ =3.792\times { 10 }^{ -16 }m$

A particle of mass m moves around the origin in a potential

$\displaystyle\dfrac{1}{2}mw^2r^2$ , where r is the distance from the origin. Applying the Bohr model in this case, the radius of the particle in its

$n^{th}$ orbit in terms of

$a=\sqrt{h/(2\pi m\omega)}$ is?

0%
$a\sqrt{n}$
0%
$an$
0%
$an^2$
0%
$an\sqrt{n}$

Explanation

The force at a distance r is

$F =- \frac{dV}{dr}=me^2r$

If r is the radius of the n-th orbit, centripetal force is equated to F.

$\frac{mv^2}{r} = me^2r$ .........(1)

Using Bohr's quantization,

$mvr = \frac{nh}{2\pi}$ ....(2)

Solving (1) and (2) by eliminating v, we get

$r = a\sqrt{n}$

The angular momentum of an electron in a hydrogen atom is proportional to(where n is principle quantum number).

0%
$n$
0%
$n^2$
0%
$n^3$
0%
$\sqrt{n}$

Explanation

According to Bohr's model
The angular momentum of orbiting electron =L =mvr =

$\dfrac{nh}{2 \pi}$
Therefore,

$L \propto n$

When a hydrogen atom is raised from the ground state to an excited state.

0%
Both K.E. and P.E. increase
0%
Both K.E. and P.E. decrease
0%
The P.E. decreases and K.E. increases
0%
The P.E. increases and K.E. decreases

An electron with kinetic energy E collides with a hydrogen atom in the ground state. The collision will be elastic.

0%
for all values of E.
0%
for $E < 10.2$ eV.
0%
for $10.2$ eV $<$ E $< 13.6$ eV only.
0%
for $0 <$ E $< 3.4$ eV only.

Explanation

$\begin{array}{l}\text { Energy }=E \\\text { Energy at ground} \text { State for hydrogen like atom}=-13.6\frac{z^{2}}{n^{2}}=V \\\text { }\text {For hydrogen atom, } z=1 \\n=1\end{array}$

$E_{1}=-13.6 \times \frac{1^2}{1^{2}}=-13.6 \mathrm{eV}$

$\text{for 2nd shell}$

$\begin{aligned}E &=-13.6 \times\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right) \\&=-10.2 \mathrm{eV}\end{aligned}$

$10.2<E<13.6$

$\therefore \text{Opion C is correct}$

2005943_784769_ans_baf5f8317baa48d2b05f5b48130ecaa9.PNG

A single electron orbits around a stationary nucleus of charge

$+Ze$ , where

$Z$ is a constant and

$e$ is the magnitude of the electronic charge. It required

$47.2\ eV$ to excite the electron from the second Bohr orbit to the third Bohr orbit. Find the value of

$Z$ .

0%
$z = 5$ .
0%
$z = 6$ .
0%
$z = 8$ .
0%
$z = 2$ .

In in nature there may not be an element for which the principal quantum number

$n> 4$ , then the total possible number of elements will be

0%
$60$
0%
$32$
0%
$4$
0%
$64$

Explanation

It is based on electronic configuration

$1s^22s^22p^63s^23p^64s^24p^64d^{10}4f^{14}$

Since maximum number of electrons which can be arranged is 60, thus there are 60 elements can be exists in nature if principle quantum number >4 were not allowed.

1003115_1034313_ans_35047d361cd24b39bba54aded58a08ef.png

In Bohr's model of the hydrogen atom, the ratio between the period of revolution of an electron in the orbit of

$n=1$ to the period of revolution of the electron in the orbit

$n=2$ is?

0%
$2:1$
0%
$1:2$
0%
$1:4$
0%
$1:8$

Explanation

$\textbf{Hint:}$ Use the time period formula and Bohr's velocity and radius formula.

$\textbf{Step 1: Writing the formulae,}$

Time period of a revolution is given by,

$T=\dfrac { 2\pi r }{ V }$

and we know that radius, r is given by

$r=\dfrac { { n }^{ 2 }{ h }^{ 2 } }{ 4{ \pi }^{ 2 }mkZ{ e }^{ 2 } }$
Similarly, velocity v is given by

$V=\dfrac { 2\pi kZ{ e }^{ 2 } }{ nh }$

$\textbf{Step 2:Substituting r and v in formula of time period}$

$T=\dfrac{ n^3\,h^3}{4\,\pi^2k^2Z^2e^3m}$

i.e

$T\propto n^3$

$\dfrac{T_1}{T_2}=\left(\dfrac{n_1}{n_2}\right)^3$

$\dfrac{T_1}{T_2}=\left(\dfrac{1}{2}\right)^3$

$\dfrac{T_1}{T_2}=\left(\dfrac{1}{8}\right)$

$\textbf{Option D is correct.}$

Energy levels I, II, III of a certain atom correspond to increasing values of energy i.e.,

$E_I < E_{II} < E_{III}$ . If

$\lambda_1$ ,

$\lambda_2$ ,

$\lambda_3$ be the wavelengths of radiations corresponding to the transitions III to II, II to I and III to I respectively, which of the following relations is correct ?

0%
$\lambda_1 = \lambda_2 + \lambda_3$
0%
$\lambda_3 = \frac {\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$
0%
$\lambda_1 + \lambda_2 + \lambda_3 = 0$
0%
$\lambda \frac{2}{3} = \lambda \frac{2}{1} + \lambda \frac{2}{2}$

Explanation

The sum of the energies for transitions III to II and II to I is equal to the energy for transition III to I. In other words, the energies and frequencies are additive whereas wavelength is inversely proportional to the frequency. So the reciprocals of the wavelengths are additive.

Thus,

$\cfrac{1}{\lambda_3}=\cfrac{1}{\lambda_2}+\cfrac{1}{\lambda_1}$

$\lambda_3=\cfrac{\lambda_1\lambda_1}{\lambda_1+\lambda_2}$

In Bohr's model of hydrogen atom, radius of the first orbit of an electron is

$r_0$ . Then, radius of the third orbit is?

0%
$\displaystyle\frac{r_0}{9}$
0%
$r_0$
0%
$3r_0$
0%
$9r_0$

Explanation

Radius of the nth orbit of an electron is given by

$r_n = r_o\dfrac{n^2}{Z}$
For hydrogen atom,

$Z = 1$

$\implies \ r_n = r_o n^2$
So, for third orbit

$n =3$
Thus radius of third orbit

$r_3 = r_o \times 3^2 = 9r_o$

If one takes into account finite mass of the proton, the correction to the binding energy of the hydrogen atom is approximately _____.(mass of proton

$1.60\times 10^{-27}\ kg$ , mass of electron

$=9.10\times 10^{-31}\ kg$ ).

0%
$0.06\%$
0%
$0.0006\%$
0%
$0.02\%$
0%
$0.00\%$

In the geiger-marsden experiment what percentage of the

$\alpha$ particles were deflected by an angle more than

$1^o$ .

0%
$< 0.15 %$
0%
$< 15 %$
0%
$> 99.85 %$
0%
$100 %$

Explanation

The Geiger-Marsden experiment, also called as the Rutherford's gold foil experiment were a landmark series of experiments by which scientists discovered that every atom contains a nucleus where all of its positive charge and most of its mass are concentrated. They deduced this by measuring how an alpha particle beam is scattered when it strikes a thin metal foil. In this experiment they observed that only less than

$15$ % of th alpha partils

CBSE Questions for Class 12 Medical Physics Atoms Quiz 8 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Atoms Quiz 8 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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