MCQ Questions for Class 12 Medical Physics Atoms Quiz 10

When

$_{90}Th^{238}$ changes into

$_{83}Bi^{222}$ , then the number of emitted

$\alpha$ and

$\beta$ particles are ?

0%
$8\alpha,7\beta$
0%
$4\alpha,7\beta$
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$4\alpha,4\beta$
0%
$4\alpha,1\beta$

Explanation

$\begin{array}{l}\text { balancing mass no. } \\\qquad 238=222+4 x+0 \\\qquad \begin{array}{l}4 x=16 \\\text { which means } 4 \alpha \text { particle released }\end{array}\end{array}$

$\begin{array}{l}\text { balaneing atomie no. } \\\qquad \begin{aligned}90 &=83+2 x+y(-1) \\7 &=8-y \\y &=1 \\\text { So, } & \text { 1 } \beta \text { particle released } \\4 \alpha, 1 \beta & \text { particle released }\end{aligned}\end{array}$

1995383_1140080_ans_6ae33c25f1164121a1d600ebcb71f90d.jpg

The quantum number corresponding to orbit of diameter 0.0001 mm in hydrogen atom will be nearly (Given that the radius of orbit with n = 1 is

$0.51\times 10^{-10}$ ):

0%
39
0%
31
0%
9
0%
49

Explanation

Radius of orbit

$=\dfrac{0.0001}{2}=5\times 10^{-5}mm=5\times 10^{-8}m$

Radius of

$n^{th}$ orbit is given by :- [from Bohr's model]

$r_n=r_0n^2$

where

$r_n=$ radius of

$1st$ orbit

$=0.51\times 10^{-10}m$

n: Quantum number of orbit

$\Rightarrow n^2=\dfrac{r_n}{r_0}=\dfrac{5\times 20^{-8}}{0.51\times 10^{-1}}$

$\Rightarrow n^2\simeq 2381$

So,

$n=\sqrt{2381 }\simeq49$

1082673_1172593_ans_58d1d67fbe204e36b6c2d63af2c55dee.png

It is observed that some of the spectral lines in hydrogen spectrum have wavelengths almost equal to those of the spectral lines in

${ H }e^{ + }$ ion, Out of the following the transitions in

${ H }e^{ + }$ that will make this is :

0%
$n = 3$ to $n = 1$
0%
$n = 6$ to $n = 4$
0%
$n = 5$ to $n = 3$
0%
$n = 3$ to $n = 2$

Explanation

Solution

Let those transitions in H be from

$n_{1}$ to

$n_{2}$ and in

$He^{+}$ be from

$n_{1}'$ to

$n_{2}'$ so,

In H,

$\dfrac{1}{h}= R(1)^{2}[\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}]$ ___ (i)

Where

$R\rightarrow$ Rydberg's constant.

similarly in

$He^{+}$ ,

$\dfrac{1}{h}= R(2)^{2}[\dfrac{1}{(n_{1}^{1})^{2}}-\dfrac{1}{(n_{2}^{1})^{2}}]$ ___ (ii)

From (i) and (ii),

$R(1)^{2}[\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}]= 4R[\dfrac{1}{n_{1}^{1}}-\dfrac{1}{n_{2}^{1}\,^{2}}]$

$\Rightarrow \dfrac{1}{4}[\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}]= \dfrac{1}{n_{1}^{1}\, ^2}-\dfrac{1}{n_{2}^{1}\,^{2}}$

$\Rightarrow \dfrac{1}{(2n_{1})^{2}}-\dfrac{1}{(2n_{2})^{2}}=\dfrac{1}{(n_{1}^{1})^{2}}-\dfrac{1}{(n_{2}^{1})^{2}}$

Thus

$n_{1}^{1}$ and

$n_{2}^{1}$ must be even no.

Option - B is correct.

1143185_1164493_ans_56b892a5a11647e5aab686226a6bc65b.jpg

The radius of the first Bohr orbit of a hydrogen atom is

$0.53\times 10^{-10}m$ .When an electron collides with this atom which is in its normal state, the radius of the electron orbit in the atom change to

$2.12 \times 10^{10}m$ .The value of the principal quantum number

$n$ of the state to which it is excited is:

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

Explanation

$r_n=a_0n^2$

$2.12\times 10^{-19}=90n^2$

$=0.53n^2$

$n^2=4$

$n=2$
This value of principle quantum number of excited state in

$2$

If energy required to remove one of the two electron from

$\text { He }$ atom is

$29.5 eV,$ then what is the value of energy required to convert a helium atom into

$\alpha -$ particle?

0%
$54.4eV$
0%
$83.9eV$
0%
$29.5eV$
0%
$24.9eV$

Explanation

From a He-atom, containing

$2e^{-}$ m ground state, one electron is

removed by proving 29.5 ev of energy, and remain

$He^{+}$

$He \overset{29.sev}{\rightarrow} He^{+}+e^{-}$

$He^{+}$ (single

$e^{-}$ species ) because uke Bohr H-atom having ground

state energy,

$E_{1} = -13.6\times \dfrac{z^{2}}{n^{2}}$ , for

$z = 2 (He)2n = 1$

$= -13.6\times \dfrac{z^{2}}{12} = -54.4ev$

So, an addition 54.4ev should be given to convent

$He^{+}$ to

$\alpha -$ partial (which is basically He nuclei, or

$He^{++}$ ).

$He^{+}\overset{54.4 ev}{\rightarrow} He^{++}+e^{-}$

Net energy required

$= 29.5+54.4 = 83.9ev$

1101264_1168012_ans_2a4d8b2a9250466592678be420ebb700.JPG

If an electron in a hydrogen atom has moved from

$n = 1$ to

$n = 10$ orbit, the potential energy of the system has:

0%
increased
0%
decreased
0%
remained unchanged
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become zero

Explanation

Solution

In Bohrs atomic at model of H -atom

$U = (-13.6ev) \dfrac{z^{2}}{n^{2}} = \dfrac{(-13.6ev(1)^{2})}{n^{2}}$

$\Rightarrow U = \dfrac{-13.6ev}{n^{2}}$

For

$n = 1,$

$U = -13.6ev$

For n = 10,

$U = -0.136 ev$

As it has got less negative, it

has been increased.

1145613_1166352_ans_9a53dc8c976f4f7e88c8566eb33a242c.jpg

When light emitted by a white hot solid is passed through a sodium flame, the spectrum of the emergent light will show

0%
The $D_{1}$ and $D_{2}$ bright yellow lines of sodium
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Two dark lines in the yellow region
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All colours from violet to red
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No colours at all

The ratio of the speed of the electron in the first Bohr orbit of hydrogen and the speed of light is equal to then (where e, h and c have their usual meanings)

0%
$\dfrac{e^{12}}{2C\epsilon^{h}}$
0%
$\dfrac{e^{2}}{2C\epsilon^{h}}$
0%
$\dfrac{e^{3}}{2C\epsilon^{h}}$
0%
$\dfrac{e^{4}}{2C\epsilon^{h}}$

Explanation

reed of an electron in

$n^{th}$ orbit (for Hydrogen- like system) is given by :

$v= \dfrac{KZl^{2}}{nh} [where\ K: \dfrac{1}{4 \pi \epsilon _{0}} \times h= \dfrac{h}{2\pi} ]$

For first orbit of Hydrogen,

$n=1\ k\ z=1$

$\Rightarrow v_{1}= \dfrac{2 \pi e^{2}}{4 \pi \epsilon_{0} h}=\dfrac{e^{2}}{2 \epsilon_{0} h}.$

Now, velocity of speed,

$=c$ , so Ratio :

$\Rightarrow \dfrac{v_{1}}{c}=\dfrac{e^{2}}{2c \epsilon_{0} h }.$

The ratio of the velocity of an electron in the first Bohr's orbit of the hydrogen atom and the velocity of light is:

0%
$1:100$
0%
$1:137$
0%
$1:1000$
0%
$1:10$

Explanation

In hydrogen atom the speed of electron is $v=\dfrac{{{e}^{2}}}{2{{\varepsilon }_{0}}nh}........(1)$

Where, e is the charge of electron

n = 1 as first orbit

Dividing equation (1) by c

$\dfrac{v}{c}=\dfrac{{{e}^{2}}}{2{{\varepsilon }_{0}}ch}$

$\dfrac{v}{c}=\dfrac{{{(1.6\times {{10}^{-19}})}^{2}}}{2\times 8.85\times {{10}^{-12}}\times 3\times {{10}^{8}}\times 6.6\times {{10}^{-34}}}$

$\dfrac{v}{c}=\dfrac{1}{137}$

Ratio is 1 : 137

How many times does the electron go round the first Bohr orbit in a second?

0%
$6.57 \times 10 ^ { 5 }$
0%
$6.57 \times 10 ^ { 10 }$
0%
$6.57 \times 10 ^ { 13 }$
0%
$6.57 \times 10 ^ { 15 }$

Explanation

Frequency

$(\nu) = \dfrac{v}{2\pi r}$

where, v is velocity of electron and r is radius of first Bohr orbit.

$\Rightarrow \nu = \dfrac{2.2 \times 10^6}{2 \pi (0.5287 \times 10^{-10} m) } = 6.57 \times 10^{15}$

Therefore, D is correct option.

Photoelectric effect was successfully explained by

0%
Planck
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Hallwash
0%
Hertz
0%
Einstein

Explanation

Einstein successfully explained the photoelectric effect using the assumption

$E=h\nu$ . This assumption is given by Planck. Einstein showed that not only light is quantized but also atomic vibrations are quantized. He was able to explain this result by assuming that the oscillations of atoms are quantized according to

$E=nh\nu$ .

Which of the following postulates of the Bohr model led to the quantization of energy of the hydrogen atom?

0%
The electron goes around the nucleus in circular orbits.
0%
The angular momentum of the electron can only be an integral multiple of $h/2\pi$ .
0%
The magnitude of the linear momentum of the electron is quantized
0%
Quantization of energy is itself a postulate of the Bohr model.

Explanation

$\begin{array}{l}\text { Quantization of of energy of the } \\\text { hydrogen atom is lead by } \\\text { angular momentum of the electron can } \\\text { only be an integral multiple of } \\h / 2 \pi.\end{array}$

a diatomic has a moment of inertia.By Bohr's quantization condition its rotational energy in the

$n^{th}$ level ( n= 0 is not allowed ) is :

0%
$\frac{1}{n^{2}}\left ( \frac{h^{2}}{8\pi ^{2}I} \right )$
0%
$\frac{1}{n}\left ( \frac{h^{2}}{8\pi ^{2}I} \right )$
0%
$n\left ( \frac{h^{2}}{8\pi ^{2}I} \right )$
0%
$n^{2}\left ( \frac{h^{2}}{8\pi ^{2}I} \right )$

Explanation

$\begin{array}{l}\qquad V=R \omega \text { , } \omega \text { is Angular velocity. } \\\text { Moment of inertia, } I=m R^{2} \\\text { Bohr's Angular momentum, } \\\qquad \begin{array}{l}m V R=\frac{n h}{2\pi} \\m \omega R^{2}=\frac{n h}{2 \pi}\end{array}\end{array}$

$\begin{array}{l}\qquad w=\frac{n h}{2 \pi m R^{2}} \\\text { Rotational energy } \\\qquad \begin{aligned}R \cdot E &=\frac{1}{2} I\omega^{2} \\&=\frac{1}{2} m R^{2} \times \frac{n^{2} h^{2}}{4\pi^{2}\left(m R^{2}\right)^{2}} \\R \cdot E&=n^{2}\left(\frac{n^{2}}{8 \pi^{2} m Q^{2}}\right) \\\therefore & R \cdot E=n^{2}\left(\frac{h^{2}}{8 \pi^{2} I}\right)\end{aligned}\end{array}$

$U^{236}$ sample of mass 1.0 g emits alpha particles at the rate

$1.24 \times 10^{4}$ particles per second. (NA =

$6.023 \times 10^{23}$ )

0%
The half life of this nuclide is $4.5 \times 10^{9}$ years
0%
The half life of this nuclide is $9 \times 1 0^{9}$ years
0%
The activity of the prepared sample is $2.48 \times 10^{4}$ particles/sec
0%
The activity of the prepared sample is $1.24 \times 10^{4}$ particles/sec.

Explanation

$\begin{array}{l}\text { No. of nucleus initially present is }\\N_{0}=\dfrac{1}{238} \times 6.023 \times 10^{23} \\N_{0}=2.531 \times 10^{21} \\\text { Activity is } \quad A=1.24 \times10^{4}\end{array}$

$\begin{array}{l}\lambda=\dfrac{A}{N_{0}}=4.90 \times 10^{-18}\\\therefore \quad T_{1 / 2}=\dfrac{\ln 2}{x}=4.49 \times 10^{9}\text { years }\end{array}$

Hence, option

$A$ and

$D$ are correct.

The orbital angular moments of an electron is

$\sqrt 3 \cfrac h {\pi}$ .Which of the following may be the permissible value of angular momentum of this electron revolving in an unknown orbit.

0%
$\cfrac h {\pi}$
0%
$\cfrac h {2\pi}$
0%
$\cfrac {3h} {2\pi}$
0%
$\cfrac {2h} {\pi}$

Explanation

$\begin{array}{l}\text { Given: orbital angular momentum, }\\\qquad\begin{aligned}\sqrt{l(l+1)} \frac{h}{2 \pi}=\sqrt{3} \frac{h}{\pi} \\\frac{l(l+1)}{4} &=3 \\\ell (l+1) &=3(3+1) \\\therefore & l=3\end{aligned}\end{array}$

$\begin{array}{l}\text { since, } l=3 \\\text { permissible values of } n \text { are } 4,5,6, \ldots . \\\text { Angular momentum can be }\frac{2 h}{\pi} \text { , } \frac{5 h}{2 \pi}, \frac{3 h}{\pi} \text { , } \ldots \\\therefore \quad \frac{2 h}{\pi}\end{array}$

In a hydrogen atom the force acting on the electron is proportional to :

0%
$n^{4}$
0%
$n^{-4}$
0%
$r^{2}$
0%
$n^{-2}$

Explanation

A hydrogen atom the force acting on the election is proportional to

$r^2$ .

Hence ,option

$C$ is correct.

$2N$ molecules of an ideal gas '

$X$ ' each of mass '

$m$ ' and

$4N$ molecules of another ideal gas

$Y,$ each of mass

$3m$ are maintained at the same absolute temperature in the same vessel. The rms velocity are

$V1$ and

$V2$ respectively then (along x-coordinate axis)

$\dfrac{v_x}{v_y}$

0%
$\sqrt { 3 } : 1$
0%
$1 : \sqrt { 3 }$
0%
$3 : 1$
0%
$1 : 3$

Explanation

$\begin{array}{l}\text { R.M.S velocity, Vrms } \alpha \sqrt{\frac{T}{m}} \\\qquad \begin{array}{l}T_{1}=T_{2} \\m_{1}=m \\m_{2}=3 m\end{array} \\\therefore \quad \frac{V_{1}}{V_{2}}=\sqrt{3} \\\text { Required ratio is } \sqrt{3}: 1\end{array}$

An electron tube was sealed off during manufacture at a pressure of

$1.2 \times 10^{-7}$ mm of mercury at

$27^oC$ . Its volume is

$100 cm^3$ . The number of molecules that remain in the tube is

0%
$2 \times 10^{16}$
0%
$3 \times 10^{15}$
0%
$3.86 \times 10^{11}$
0%
$5 \times 10^{11}$

Explanation

$P=nkT\\ n=\cfrac { P }{ kT } =\cfrac { hdg }{ kT }$

Number of molecules in the electron tube is

$=n\times V=\cfrac { hdgV }{ kT }$

$\Longrightarrow n=\cfrac { hdg }{ kT } =\cfrac { 1.2\times { 10 }^{ -10 }\times 13600\times 9.8\times 100\times { 10 }^{ -6 } }{ 1.38\times { 10 }^{ -23 }\times 300 } =3.86\times { 10 }^{ 11 }$

The ratio of the orbit of the

$1st$ three radii in an atom of hydrogen is

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

Explanation

$\begin{array}{l}\text { Radii of nth orbit in an hydrogen } \\\text { atom is proportional to n }^{2} \\\therefore \text { Required ratio is } \\\qquad \begin{array}{l}(1)^{2}:(2)^{2}:(3)^{2} \\=1: 4: 9\end{array}\end{array}$

Ratio of series limit of Brackett and Pfund series of hydrogen ato spectrum is

0%
$\dfrac{1}{4}$
0%
$\dfrac{4}{9}$
0%
$\dfrac{9}{16}$
0%
$\dfrac{16}{25}$

Explanation

$\begin{array}{c}{\text { Bracket series }} \text { n }_{1}=3, n_{2}=4,5,\ldots \\\dfrac{1}{\text { series limit }}\propto \left[\dfrac{1}{3^{2}}-\dfrac{1}{n_2^2}\right] \\\propto \dfrac{1}{9}\end{array}$

$\begin{array}{l}\text { Pfund series } \quad n_{1}=4, n_{2}=5,6,\ldots \\\dfrac{1}{\text { series limit }}\propto \left[\dfrac{1}{4^{2}}-\dfrac{1}{(n_2)^{2}}\right] \propto \dfrac{1}{16} \\\text { Ratio of series limit }=9/16\end{array}$

Orbital acceleration of electron in first orbit of hydrogen atom is:

0%
$\cfrac{h}{2\pi^2m^2r^3}$
0%
$\cfrac{h^2}{2\pi^2m^2r^3}$
0%
$\cfrac{h^2}{4\pi^2m^2r^3}$
0%
$\cfrac{h^2}{4\pi^2m^2r^2}$

Explanation

Orbital Angular momentum

$mvr=\dfrac{nh}{2\pi }$ n=1

$mvr=\dfrac{h}{2\pi }$

$v=\dfrac{h}{2\pi mr}$

$v^{2}=\left ( \dfrac{h}{2\pi mr} \right )^{2}$

$a=\dfrac{V^{2}}{r}=\dfrac{h^{2}}{4\pi ^{2}m^{2}r^{3}}$

Equivalent electric current created by electron of a H-atom in its ground using Bohr's model is nearly

0%
$2.14\times {10}^{-35}A$
0%
$4.48\times {10}^{-30}A$
0%
$24\times {10}^{-30}A$
0%
$70\times {10}^{-35}A$

Explanation

$\begin{array}{l}\text { Let, radius of } 1^{\text {St }} \text { orbit Atom be "a_o" } \\\text { velocity of } e^{-} \text { in } 1^{\text {st }} \text { orbit is } v \\\text { Time period, }\left[T=\dfrac{2 \pi \cdot a_{0}}{v}]--(1)\right] \end{array}$

$\begin{array}{l}\text { Electric current, } i=\dfrac{q}{T} \\\text { From } \\\qquad i=\dfrac{q v}{2 \pi a_{0}}\end{array}$

$\begin{array}{l}\text { By keeping approximate experimental } \\\text { values in above expression, we get } \\\qquad i=4.489 \times 10^{-30} A\end{array}$

Find the frequency of revolution of the electron in the first stationary orbit of H-atom

0%
$8\times 10^{14}Hz$
0%
$8.6\times 10^{10}Hz$
0%
$8.6\times 10^{-10}Hz$
0%
$8.12\times 10^{16}Hz$

Explanation

$frame=\dfrac{1}{Time\ period}$

Time perid

$=\dfrac{Total \ distance \ coverd}{velocity}=\dfrac{2\pi r}{v}$

frequency

$=\dfrac{v}{2\pi r}$

The velocity

$v_2$ and radius

$r_2$ of the

$2^{nd}$ both orbit

$r_n=(0.53\times 10^{-10}n^2/Z)\ m$ ,

$Z=1\ and\ n=2$

$r_2=0.53\times10^{-10}(2^2)\ =2.12\times 10^{-10}m$

$v_n=2.165\times 10^{6}\left(2/n\right)m/s$

$v_2=2.165\times 10^{6}\left(1/2 \right)=1.082\times 10^6m/s$

frequency

$=\dfrac{v_2}{2\pi r_2}=\dfrac{1.082\times 10^6}{2(x)(2.12\times 10^{-10})}$

$f=8.13\times 10^{16}Hz$

In hydrogen atom, the electron is making

$6.6\times 10^{15}rev/s$ around the nucleus in an orbit of radius

$0.528A^0$ . The equivalent magnetic dipole moment is :

0%
$1\times 10^{-15}Am^2$
0%
$1\times 10^{-10}Am^2$
0%
$1\times 10^{-23}Am^2$
0%
$1\times 10^{-27}Am^2$

Alpha particles are projected towards fixed at a nucleus. Which of the paths shown in figure, is not possible

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

The Bohr radius of the fifth electron of phosphorus (atomic number = 15) acting as dopant in silicon (relative dielectric constant = 12) is

0%
10.6 $\ { A^0 }$
0%
0.53 $\ { { A ^0} }$
0%
21.2 ${ A^0 }$
0%
None of these

Explanation

${r_n} = { \in _r}\left( {\frac{{{n^2}}}{Z}} \right)$

${a_0} = 12 \times \frac{{{{\left( 5 \right)}^2}}}{{15}} \times 0.53 = 10.6\mathop A\limits^0$

Hence,

option

$(A)$ is correct answer.

The radius of the orbit of an electron in a Hydrogen-like atom is

$3{ a }_{ 0 }$ , where

${ a }_{ 0 }$ is the Bohr radius. Its orbital angular momentum is

$\frac { 3h }{ 2\pi }$ . It is given that h is Planck's constant and R is Rydberg constant. The possible wavelength, when the atom de-excites is :

0%
$\dfrac { 4 }{ 5R }$
0%
$\dfrac { 4 }{ 9R }$
0%
$\dfrac { 1 }{ 2R }$
0%
$\dfrac { 9 }{ 32R }$

Explanation

solution:

given ;

$4.5a_0 = a_0 \dfrac{n^2}{Z}----(i)$

$\dfrac{nh}{2\pi} = \dfrac{3h}{2\pi} ----(ii)$

so , n= 3 and z=2

so , possible wavelength are:

$\dfrac{1}{\lambda _1}= RZ^2[\dfrac{1}{1^2} - \dfrac{1}{3^2}] \Rightarrow \lambda _1= \dfrac{9}{32R}$

hence the correct option:D

What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series ?

0%
4 : 1
0%
4 : 9
0%
4 : 3
0%
5 : 9

Explanation

For transition

$n_2 \rightarrow n_1$

the wavelength is given by

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

For shortest wavelength in Lyman series put

$n_1=1$ and

$n_2 \rightarrow \infty$

and for shortest wavelength in Balmer series put

$n_1=2$ and

$n_2 \rightarrow \infty$

we get

$\dfrac{1}{\lambda _L}=RZ^2$

and

$\dfrac{1}{\lambda _B}=RZ^2(\dfrac{1}{4})$

ratio will be

$\dfrac{\lambda _B}{\lambda _L}=\dfrac{4}{1}$

Option A is correct.

The number of bond pair of electrons and lone pair of electrons in

$O_2$ molecules respectively are

0%
2, 2
0%
2 ,1
0%
4, 2
0%
2, 4

In a cubical vessel $1m \times 1m\times 1m$ the gas molecules of diameter $1.7 \times {10^{ - 8}}\;{\text{cm}}$ are at a temperature 300 K and a pressure of ${10^{ - 4}}\;{\text{mm}}$ mercury. The mean free path of the gas molecule is

0%
1 meter
0%
4 meter
0%
2.42 meter
0%
1 cm

Explanation

The mean free path or average distance between collison for a gas molecule may be estimated from KTG(Kinetic Theory of Gases).

From Serway's approach:-

$(Mean\quad free\quad path)\lambda=\cfrac{1}{\pi d^2n_v}$

$d\rightarrow$ diameter of molecule

$n_v\rightarrow$ No. of molecules per unit volume

From idle gas law,

$n_v=\cfrac{nN_A}{V}\Rightarrow n_v=\cfrac{N_AP}{RT}$

So, from

$(i)$

$\lambda=\cfrac{RT}{\sqrt{2}\pi d^2N_A P}$

A/Q we have

$d=1.7\times 10^{-8}cm=17\times10^{-10}m\\P=10^{-4}mmHg\\T=300K\\ \lambda=\cfrac{8.314\times 300}{\sqrt2 \times3.14\times (1.7\times10^{-10})^2}(6.022\times10^{23}(10^{-4}))\\ \quad=\cfrac{8.314\times 3\times10^3}{1.414\times3.14\times2.89\times6.022}\\ \quad=2.4201m$

The figure indicates the energy level diagram of an atoms and the origin of six spectrum lines in emission (e.g line no.

$5$ arises from the transition from level

$B$ to

$A$ ) Which of the following spectral lines will occur in the absorption spectrum?

0%
$4,5,6$
0%
$1,2,3,4,5,6$
0%
$1.2.3$
0%
$1,4,6$

Explanation

The absorption lines are obtained when the electron jumps from ground state

$(n = 1)$ to the higher energy states. Thus only

$1, 2$ and

$3$ lines will be obtained.

Frequencies higher than

$10 MHz$ are found not to be reflected by the particles present in the ionosphere then what is the maximum electron density of the ionosphere?

0%
$\dfrac{{{{10}^{14}}}}{81}e{m^{-3}}$
0%
${10^{14}}e{m^{ - 3}}$
0%
${10^{4}}e{m^{ - 3}}$
0%
$\dfrac{{{{10}^{14}}}}{7}e{m^{ - 3}}$

Explanation

Given,

Critical frequency

$(f_c)=10MHz=10^7 Hz$

Let maximum electrons density in the ionosphere be

$N_{max}$

The relation between the critical frequency and the maximum electron density of the ionosphere is given by the expression:

$f_c=9\times(N_{max})^{1/2}$

$N_{max}=\dfrac{f_c^2}{81}$

$N_{max}=\dfrac{10^{14}}{81}$

So, the answer will be option

$(A)$

What will be relation between first bohr radius of H- atom and D-atom

0%
$r _ { H } = r _ { D }$
0%
$r _ { H } > r _ { D }$
0%
$r _ { \mathrm { H } } < \mathrm { r } _ { \mathrm { D } }$
0%
Can't predict

Which of the following is quantised according to Bohr's theory of hydrogen atom

0%
Linear momentum of electron
0%
Angular momentum of electron
0%
Linear velocity of electron
0%
Angular velocity of electron

In a beryllium atom, if

$a_{0}$ be the radius of the first orbit, then the radius of the second will be in general.

0%
$na_{0}$
0%
$a_{0}$
0%
$n^{2}a_{0}$
0%
$\dfrac {a_{0}}{n^{2}}$

Explanation

$F\propto n^2$

$r=\dfrac{n^2h^2\epsilon _0}{\pi mZe^2}$

$r_1=\dfrac{h^2\epsilon _0 }{\pi mZe^2}=a_0$

$r_n=\dfrac{n^2h^2\epsilon_0}{\pi mZe^2}$

$r_n=a_0n^2$

The colour which corresponds to

$\mathrm { H } _ { \mathrm { B } } ( 4 \rightarrow 2 )$ linein the Balmer series of hydrogen spectra is:

0%
Green
0%
Red
0%
Violet
0%
Blue

On a particular day, the maximum frequency reflected from the ionosphere is 10 MHz. On another day, it was found to increase to 11 MHz. Calculate the ratio of maximum electron density of the ionosphere on the two days.

0%
$1.21$
0%
$0.82$
0%
$0.50$
0%
$0.25$

Explanation

Max. freq. of a

${ 1 }^{ st }$ day

$=10{ MH }_{ z }$

On a subseqent day

$=10{ NH }_{ 2 }$

Max electron density on

${ 1 }^{ st }$ day

${ N }_{ max }={ \left( \dfrac { 10 }{ 9 } \right) }^{ 2 }=\dfrac { 100 }{ 81 }$

Max electron density on

${ 2 }^{ nd }$ day

${ N }_{ max }={ \left( \dfrac { 11 }{ 9 } \right) }^{ 2 }=\dfrac { 121 }{ 81 }$

ratio

$=\dfrac { \dfrac { 100 }{ 81 } }{ \dfrac { 121 }{ 81 } } =\dfrac { 121 }{ 100 } =1.21$

When hydrogen atom is in its first excited level, its radius is _______ of the Bohr radius.

0%
Twice
0%
$4$ times
0%
Same
0%
Half

Explanation

When hydrogen atom is in its first excited level,

$n=2$

The radius of hydrogen atom in the first excited level is

$r=n^2r_0$

$r=(2)^2r_0=4r_0$

where,

$r_0=$ Bohr's radius.

The correct option is B.

In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is

$\lambda$ . If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be?

0%
$\dfrac{27}{20}\lambda$
0%
$\dfrac{16}{25}\lambda$
0%
$\dfrac{20}{27}\lambda$
0%
$\dfrac{25}{16}\lambda$

Explanation

For M

$\rightarrow$ L steel

$\dfrac{1}{\lambda}=K\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)=\dfrac{K\times 5}{36}$
for N

$\rightarrow$ L

$\dfrac{1}{\lambda'}=K\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)=\dfrac{K\times 3}{16}$

$\lambda'=\dfrac{20}{27}\lambda$ .

A proton and an

$\alpha$ particle having equal kinetic energy are projected in a uniform transverse electric field as shown in figure

0%
Proton trajectory is more curved
0%
$\alpha$ particle trajectory is more curve
0%
Both trajectories are equally curved but in opposite direction
0%
Both trajectories are equally curved and in same direction

Explanation

Equal kinetic energy.

But the trajectory will depend on

$acceleration$ which is

$F=\dfrac{qE}{m}$ Which is

$\dfrac{eE}{m_p}$ for proton and

$\dfrac{2eE}{4m_p}=\dfrac{eE}{2m_p}$ for

$\alpha$ particle.

$E$ is electric field ,

$q$ is charge and

$m_p$ is mass of

$one$ proton.

$\text{As they are opposite in charge so opposite direction and proton having more acceleration will show}$

$\text{ more curved trajectory. }$

Option A is correct.

Imagine an atom made up of proton and a a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength

$\lambda$ (give in term of the Rydberg constant R for the hydrogen atom) equal to

0%
9/5 R
0%
36/5 R
0%
18/5 R
0%
4/ R

Explanation

Energy is related to mass

${R_n} \propto m$

The longest wavelength λmaxλmax photon will correspond to the transition of particle from

$n=3$ to

$n=2$

$\begin{array}{l} \dfrac { 1 }{ { { \lambda _{ \max } } } } =2R\left( { \dfrac { 1 }{ { { 2^{ 2 } } } } -\dfrac { 1 }{ { { 3^{ 2 } } } } } \right) \\ { \lambda _{ \max } }=\dfrac { { 18 } }{ 5 } R \end{array}$

Hence, Option

$C$ is correct.

In Bohr series of lines of hydrogen spectrum, third line from the red end corresponds to which one of the following inner orbit jumps of electron for Bohr's orbit in atom of hydrogen

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

Explanation

We see red end means it is a part of the visible region and obviously only Balmer series corresponds to the visible region for the Balmer series

$n_1=2$ and red end means low energy or third end from this means

$n_2=-5$

$2(r)5. 3$

$2(r)4. 2$

$2(r)3. 1$

The correct answer is (D)

$5 \to 2$

The distance between

$3^{rd}$ &

$2^{nd}$ Bohr orbit

$He^{+}$ is

0%
$2.645\ A^o$
0%
$1.322\ A^o$
0%
$0.2645\ A^{o}$
0%
$None\ of\ these$

Explanation

Dis\tance between the

${ 2^{ nd } }$ and the

${ 3^{ rd } }$ orbits=distance of

${ 3^{ rd } }$ orbit-distance of second orbit from the nucleus

${ d_{ 3 } } = 0.529 \times \dfrac { n^{ 2 }}{ z } \times { 10^{ -10 } } m=0.529 \times 9 \times { 10^{ -10 } } m$

${ d_{ 2 } }=0.529\times 4\times { 10^{ -10 } }\, m \\ { d_{ 3 } }-{ d_{ 2 } }=0.529\times 5\times { 10^{ -10 } }\, m =2.645\times { 10^{ -10 } }\, m=2.645A^{ 0 }$

Hence,

option

$(A)$ is correct answer.

The ratio of the radii of the first three Bohr orbit in H atom is

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

Explanation

Atomic number

$Z$ is equal to

$1$ .

Hence the radius of

$n th$ orbit,

$r_n=0.529 n^2 A$

For first three orbits,

$n$ values are

$1,2$ and

$3$

ratio of radii of first three orbit

${r_1}:{r_2}:{r_3} = n_1^2:n_2^2:n_3^2$

$\begin{array}{l} \Rightarrow { 1^{ 2 } }:{ 2^{ 2 } }:{ 3^{ 2 } } \\ \Rightarrow 1:4:9 \end{array}$

the ground state energy of hydrogen atom is -13.6 eV . the energy of second excited state of

$He^{ + }$ ion in eV is :

0%
$-27.2$
0%
$-3.4$
0%
$-54.4$
0%
$-6.04$

Explanation

$\begin{array}{l} E=-13.6\times \dfrac { { { z^{ 2 } } } }{ { { n^{ 2 } } } } \\ =-13.6\times \dfrac { 4 }{ 9 } \\ =-6.04\, eV \end{array}$

Hence,

option

$(D)$ is correct answer.

The angular momentum of electron in 4th orbit of hydrogen atom is

0%
$\dfrac { 4h }{ \pi }$
0%
$\dfrac { 2h }{ \pi }$
0%
$\dfrac { nh }{ 2\pi }$
0%
$\dfrac { h }{ 4\pi }$

Explanation

The angular momentum of electron in

$4th$ orbit of hydrogen atom is

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

Hence, Option

$C$ is correct.

Imagine an atom made of a proton and a hypothetical particle of double the mass as that of an electron but the same charge. Apply Bohr theory to consider transitions of the hypothetical particle to the ground state . Then, the longest wavelength (in terms of Rydberge constant for hydrogen atom ) is

0%
$\dfrac { 1 } { 2 R }$
0%
$\dfrac { 5} { 3 R }$
0%
$\dfrac { 1 } { 3 R }$
0%
$\dfrac { 2 } { 3 R }$

According to Bohr's model, if the kinetic energy of an electron in

$2^{nd}$ orbit of

$He^{+}$ is

$x$ , then what should be the ionisaion energy of the electron revolving in

$3^{rd}$ orbit of

$m^{5+} ions>$

0%
$X$
0%
$4X$
0%
$x/4$
0%
$2X$

Explanation

$\begin{array}{l} X=13\cdot 6\frac { { { { \left( 2 \right) }^{ 2 } } } }{ { { { \left( 2 \right) }^{ 2 } } } } \\ =13\cdot 6 \\ X'=13\cdot \frac { { { { \left( 6 \right) }^{ 2 } } } }{ { { { \left( 3 \right) }^{ 2 } } } } \\ =13\cdot 6\times \frac { { 36 } }{ 9 } \\ =4X \\ Hence,\, option\, B\, is\, correct\, answer. \end{array}$

Two electron separated by a distance

$r$ experiences force

$F$ between them . Then force between a proton and a singly ionized helium atom separated by a distance

$2r$ is

0%
$4F$
0%
$2F$
0%
$\frac { F } {2}$
0%
$\frac { F } {4 }$

Explanation

According to Coulomb's law

$\begin{array}{l} F=\dfrac { 1 }{ { 4\pi { E_{ 0 } } } } .\dfrac { { \left( e \right) \left( e \right) } }{ { { r^{ 2 } } } } \\ { F_{ 1 } }=\dfrac { 1 }{ { 4\pi { E_{ 0 } } } } .\dfrac { { \left( e \right) \left( e \right) } }{ { { r^{ 2 } } } } \end{array}$

Charge on proton is equal to charge on electron and charge on singly Ionized

atom separated by distance

$'2r'$

$\begin{array}{l} { F_{ 2 } }=\dfrac { 1 }{ { 4\pi { E_{ 0 } } } } .\frac { { \left( { e\lambda c } \right) } }{ { { { \left( { 2r } \right) }^{ 2 } } } } \\ =\dfrac { 1 }{ { 4\pi { E_{ 0 } } } } .\dfrac { { { e^{ 2 } } } }{ { 4{ r^{ 2 } } } } \\ \dfrac { { { F_{ 1 } } } }{ { { F_{ 2 } } } } =\dfrac { F }{ 4 } \end{array}$

Option D.

Hydrogen atom in ground state is excited by a monochromatic radiation of

$\lambda =975\mathring { A }$ Number of spectral lines in the resulting spectrum emitted will be

0%
$3$
0%
$2$
0%
$6$
0%
$10$

CBSE Questions for Class 12 Medical Physics Atoms Quiz 10 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Atoms Quiz 10 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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