MCQ Questions for Class 12 Medical Physics Atoms Quiz 9

What is alpha-particle?

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The nucleus of hydrogen atom
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The nucleus of deuterium
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A fast moving proton
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The nucleus of helium atom

Explanation

$\begin{array}{l}\text { Alpha particle is the nucleus of a } \\\text { helium atom } ^4_2 \mathrm{He} \text { . }\end{array}$

If the first line of Lyman series has a wavelength

$1215.4\mathring { A }$ , the first line of Balmer series is approximately

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$4864\mathring { A }$
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$1025.5\mathring { A }$
0%
$6563\mathring { A }$
0%
$6400\mathring { A }$

Explanation

Wavelength of first line in Lyman series

$\lambda_L =\dfrac{3}{4R}$

$\therefore$

$\dfrac{4}{3R} = 1215.4A^o$
We get

$R = \dfrac{4}{3\times 1215.4}$
Wavelength of first line in Balmer series

$\lambda_B =\dfrac{36}{5R}$

$\implies \ \lambda_B = \dfrac{36\times 3\times 1215.4}{5\times 4} = 6563A^o$

The wavelength of spectral line coming from a distant star shifts from

$600nm$ to

$600.1 nm$ . The velocity of the star relative to earth is

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$50 km s^{-1}$
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$100 km s^{-1}$
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$25 km s^{-1}$
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$200 km s^{-1}$

Explanation

Here,

$\lambda =600 nm$

$\Delta \lambda=600.1nm-600nm=0.1nm$

$\dfrac{v_s}{c}=\dfrac {\Delta\lambda}{\lambda}$

$\therefore v_s=\dfrac {\Delta\lambda}{\lambda}c=\dfrac{0.1nm}{600nm}\times 3\times 10^8 m s^{-1}$

$V_s=50 \times 10^3 m s^{-1}$

$V_s=50 km s^{-1}$

In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by

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increasing the potential difference between the anode and filament
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increasing the filament current
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decreasing the filament current
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decreasing the potential difference between the anode and filament

Cathode rays were discovered by

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Maxwell Clerk James
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Heinrich Hertz
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William Crookes
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J. J. Thomson

Photons have properties similar to that of.

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Both particles and waves
0%
Particles
0%
Waves
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Neither particles nor waves

Number of spectral line in hydrogen atom is

0%
$6$
0%
$8$
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$15$
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None of the above

The photon energy in units of eV for electromagnetic waves of wavelength

$2 cm$ is:

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$2.5 10^{-19}$
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$5.2 10^{-16}$
0%
$3.2 10^{-16}$
0%
$6.2 10^{-5}$

Explanation

Given,

$\lambda=2cm=0.02m$

$h=6.62\times 10^{-34} kgm^2/s$

$c=3\times 10^8m/s$

Photon energy,

$E=\dfrac{hc}{\lambda}$

$E=\dfrac{6.62\times 10^{-34}\times 3\times 10^8}{0.02}$

$E=993\times 10^{-26}J$

Photon energy in unit of

$eV$ ,

$E=\dfrac{993\times 10^{-26}}{1.6\times 10^{-19}}eV$

$E=6.20\times 10^{-5}eV$

The ionization energy of a hydrogen like Bohr atom is

$4$ Rydbergs. (i) What is the wavelength of the radiation emitted when an electron jumps from the first excited state to the ground state? (ii) What is the radius of the first orbit for this atom?

0%
$300\overset {\circ}{A}, 2.65\times 10^{-11}m$ .
0%
$600\overset {\circ}{A}, 2.65\times 10^{-11}m$ .
0%
$300\overset {\circ}{A}, 4\times 10^{-11}m$ .
0%
$600\overset {\circ}{A}, 4\times 10^{-11}m$ .

A double ionized lithium atom is hydrogen-like with atomic number

$3$ .
(a) Find the wavelength of the radiation required to excite the electron in

$Li++$ from the first to the third Bohr orbit. (Ionization energy of the hydrogen atom equals

$13.6\ eV$ )\
(b) How many spectral lines are observed in the emission spectrum of the above excited system?

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$114.7\ A^{\circ}$ , Three lines.
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$113.7\ A^{\circ}$ , four lines.
0%
$115.7\ A^{\circ}$ , two lines.
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$116.7\ A^{\circ}$ , one lines.

Explanation

formula used

$=-13.6\dfrac{z^{2}}{n^{2}}eV$

now, for

$n=1, E_{i}=-122.4\ eV$

for

$n=3, E_{j}=-13.6\ eV$

now, photon energy required

$E_{p}=E_{j}-E_{i}$

$\dfrac{h o}{\lambda}=\left\{-13.6-(-122.4)\right\}eV$

$\lambda=\dfrac{12400\ eV}{108.8\ eV}-A$

$\lambda=113.97\ A$

spectral lines are

$3\rightarrow 2\rightarrow 1\rightarrow 0$

$\therefore$ four lines

$\therefore$ Option

$(B)113.7\ A^{o}$ , four lines is correct option

An electron with energy

$12.09$ eV strikes hydrogen atom in ground state and gives its all energy to the hydrogen atom. Therefore hydrogen atom is excited to __________ state.

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Fourth
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Third
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Second
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First

Explanation

We know that energy of Hydrogen atom in it's orbit is given by formula

$\dfrac{-13.6}{n^2} eV$ .

Initial Energy of Electron =

$-13.6 eV$ (Given that atom in i's ground state i.e n=1)

Energy after striking =

$(-13.6 + 12.09) eV = -1.51 eV$

$\Rightarrow$

$\dfrac{-13.6}{n^2} eV = 1.51eV$

$\Rightarrow$

$n^2= 9$

$\Rightarrow$

$n=3.$

So, hydrogen atom is excited to third state.

Therefore, B is correct option.

The radius of second orbit in an atom of hydrogen is R. What is the radius in third orbit.

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$3R$
0%
$2.25R$
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$9R$
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$\dfrac{R}{3}$

Explanation

$r_n=0.529\dfrac{n^{2}}{Z} \overset{o}{A}$ where

$r_n$ is radius of

$n^{th}$ orbit.

$n=$ number of the orbit.

$Z=$ atomic number.

Given:

$Z=1$ (Hydrogen orbit)

$R=0.529\times \dfrac{(2)^{2}}{1}=4\times 0.529$

Now let radius of

$3^{th}$ orbit be x.

$x=0.529\dfrac{n^{2}}{Z} \overset{o}{A}=0.529\dfrac{(3)^{2}}{Z}\overset{o}{A}=0.529\times \dfrac{9}{1} \overset{o}{A}=0.529\times\dfrac{4}{4} \times \dfrac{9}{1} \overset{o}{A}=R\times \dfrac{9}{4}=2.25R$

Hence the correct option is (B).

In the Bohr model of the hydrogen atom.

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The radius of the $n^{th}$ orbit is proportional to $n^{2}$
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The total energy of the electron in the $n^{th}$ orbit is inversely proportional to $'n'$
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The angular momentum of the electron in an orbit is an integral multiple of $h/2\pi$
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The magnitude of the potential energy of the electron in any orbit is greater than its kinetic energy

Explanation

Radius of Bohr's orbit of hydrogen atom is given by

$r=\cfrac{n^2h^2}{4\pi^2 mKze^2}$ or

$r=(0.59 \mathring { A } )\cfrac { n^{ 2 } }{ z }$

So from expression we found

$r \propto n^2$

And according to Bohr's theory

Angular momentum of elctron in an orbit will be Integral multiple of

$(h/2\pi)$

Magnitude of potential energy is twice of kinetic energy of electron in an orbit

$\left| P.E \right| =2\left| K.E \right|$

$K.E=(13.6ev)\cfrac{(z^2)}{(n^2)}$

If element with principal quantum number

$n > 4$ were not allowed in nature, the number of possible elements would be

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

Explanation

If the element with principal quantum

number

$n>4$

when

$n=4$

The formula becomes

$=2(4)^2\\=16\times2\\=32$

When

$n=5$

The formula becomes

$=2(5)^2$

Hence the number of elements would

$=60$

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 wavelength of the electromagnetic radiation required to remove the electron from the first Bohr orbit.

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$36.4$
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$53$
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$40$
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$66$

The number of waves in the third orbit of H atom is

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1
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2
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4
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3

The total energy of in the electron in the hydrogen atom in the ground state is

$-13.6\ eV$ . Which of the following is its kinetic energy in the first exited state?

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$13.6 \ eV$
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$6.8 \ eV$
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$3.4 \ eV$
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$1.825 \ eV$

Explanation

Energy nth orbit of hydrogen atom is given by,

$E_n=\dfrac{-13.6}{n^2}e.V$

For ground state,

$n=1$

$\therefore E_l=\dfrac{-13.6}{l^2}=-13.6eV$

for first excited state,

$n=2$

$\therefore E_2=\dfrac{-136.6}{2^2}=\dfrac{-136.6}{4}=-3.4eV$

Kinetic energy of an electron in the first excited state is,

$K.E=-E_2$

$\Rightarrow K.E=3.4eV$

The radius of the first Bohr orbit is

$a_{0}$ . The

$n^{th}$ orbit has a radius:

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$na_{0}$
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$a_{0}/n$
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$n^{2}a_{0}$
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$a_{0}/n^{2}$

Explanation

Radius of

$n^{th}$ orbital

$r_n=\dfrac{\epsilon_0n^{2}h^{2}}{\pi mZe^2}$

Wherein

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

$=\dfrac{\epsilon_0h^{2}}{\pi me^2}=0.529$

$r=\dfrac{n^2}{Z}a_0$

For

$Z=1$

$r=n^{2}a_0$

The average current due to an electron orbiting the proton in the

$n^{th}$ Bohr orbit of the hydrogen atom is

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$\propto n$
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$\propto n^{3}$
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$\propto n^{-3}$
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none of these

Explanation

We know,

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

$v_n \propto \dfrac{Z}{n}$

$\omega=\dfrac{2\pi r_n}{v_n}$

$\omega\propto\dfrac{Z}{n^3}$

$I\propto\omega\propto\dfrac{1}{n^{-3}}$

$L\propto n^{-3}$

Hence option

$\textbf C$ is correct.

The ionisation potential of a hydrogen atom is

$13.6\ volt$ . The energy required to remove an electron from the second orbit of hydrogen is:

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$3.4\ eV$
0%
$6.8\ eV$
0%
$13.6\ eV$
0%
$27.2\ eV$

Explanation

Energy required to remove electron in the

$n=2$ state

$E_n=-13.6\dfrac{Z^2}{n^2}$

$\Delta E=-13.6\left[\dfrac{1}{\infty^2}-\dfrac{1}{4}\right]$

$\Delta E=\dfrac{13.6}{4}eV$

$=3.4\ eV$

The radius of the shortest orbit in a one-electron system is

$18\ pm$ . It may be

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Hydrogen
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Deuterium
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$He^{+}$
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$Li^{2+}$

Three photons coming from emission spectra of hydrogen sample are picked up. Their energies are

$1eV, 10.2eV$ and

$1.9eV$ . These photons must come from

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a single atom
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two atoms
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three atom
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either two atoms or three atoms

The ratio of the speed of an electron in the first orbit of hydrogen atom to that in the first orbit of

$He^{+}$ is

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$1\ :\ 2$
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$2\ :\ 1$
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$1\ :\ 4$
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$4\ :\ 1$

Explanation

$\begin{array}{l}\text { From single electron model we get - }\\\qquad\begin{aligned}\frac{m v^{2}}{r} &=\frac{k z e^{2}}{r^{2}}\\\Rightarrow \quad m v^{2} r =k z e^{2}\end{aligned} \\\text {Rearranging } \rightarrow \quad v \cdot(m v r)=k z e^{2}\end{array}$

$\begin{array}{l}\text { By Bohr quantization we have }\left[\text { mvr }=\frac{n h}{2 \pi}\right] \\\text { So putting in above eq we get, } \\\qquad \begin{array}{l}v \cdot \frac{n h}{2 \pi}=k z e^{2} \\\Rightarrow v=\frac{2 \pi k z e^{2}}{n h}\end{array}\end{array}$

$\begin{array}{l}\Rightarrow \quad \frac{V_{H}}{V_{H e}}=\frac{Z_{H}}{n_{H}} \times \frac{n_{H e}}{Z_{H e}}=\frac{1}{1} \times \frac{1}{2}=\frac{1}{2} \\\Rightarrow \text { Velocity of hydrogen to velocity of Helium } \\\qquad V_{H}: V_{H e}=1: 2\end{array}$

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 value 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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$n_1=6, n_2=3$

Explanation

$frequency \propto \cfrac{z^2}{n^3}$

$\Rightarrow f=\cfrac{kz^2}{n^3}$

$\therefore f_1=\cfrac{1}{27}f_2$

$\Rightarrow \cfrac{kz_1^2}{n_1^3}=\cfrac{1}{27}\cfrac{kz_2^2}{n_2^3}\quad \left(z_1=z_2=1\right)$

$\Rightarrow \cfrac{k}{n_1^3}=\cfrac{1}{27}\cfrac{k}{n_2^3}$

$\Rightarrow n_1=3n_2$

$\therefore n_1=3$ and

$n_2=1$

The key feature Of Bohr's theory Of spectrum Of hydrogen atom is the quantization Of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy Of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.A diatomic molecule has moment Of inertia I. By Bohr's quantization condition its rotational energy in the nth level (n=0 is not allowed) is

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$\cfrac{h^2}{(n^2)(8\pi^2I)}$
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$\cfrac{h^2}{(n)(8\pi^2I)}$
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$\cfrac{n h^2}{8\pi^2I}$
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$\cfrac{n^2 h^2}{(8\pi^2I)}$

Explanation

$\begin{array}{l}\text { Given - * A Diatomic molecule- } \\\text { let, the structure of the molecule be like- }\end{array}$

$\Rightarrow I=2 m r^{2}$

$\begin{array}{l}\text { we howe Rotational encrgy- } \\\qquad E=\frac{1}{2} I \omega^{2}=m \omega^{2} r^{2}-(1) \\\text { By Bohr quantization of angular momentum } \\\qquad 2 \text { mwr}^2=\frac{n h}{2 \pi} \quad-(2) \\\text { From } e q(1) \text { and (2) we get ,}\\\qquad E=\frac{n^{2} h^{2}}{16 \pi^{2} m r^{2}}=\frac{n^{2} h^{2}}{8\pi^{2} I}\end{array}$

1994219_1068496_ans_86b7fb048fa04ad6bc709fc0c20ec337.JPG

An electron in hydrogen atom makes a transition

$n_{1}\rightarrow n_{2}$ where

$n_{1}$ and

$n_{2}$ are principal quantum numbers of the two states. Assuming Bohr's model to be valid, the time period of the electron in the initial state is eight times that in the final state. The possible values of

$n_{1}$ and

$n_{2}$ are

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$n_{1} = 6$ and $n_{2} = 2$
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$n_{1} = 8$ and $n_{2} = 1$
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$n_{1} = 8$ and $n_{2} = 2$
0%
$n_{1} = 4$ and $n_{2} = 2$

Explanation

$\begin{array}{l}\text { Given- * An electron makes transition } n_{1}\longrightarrow n_{2} \\\text { such that Time period in initial state is } \\8 \text { times the final state. }\left[T_{1}=8T_{2}\right]\end{array}$

$\text{We know that in a single $e^{-}$ model-}$

$T \propto \frac{n^{3}}{z^{2}}$

$\Rightarrow \frac{T_{1}}{T_{2}}=\frac{n_{1}^{3}}{n_{2}^{3}}=\frac{8}{1}$

$\begin{aligned}\Rightarrow \frac{n_{1}}{n_{2}}=2 \Rightarrow n_{1}=4\quad n_{2}=2 \\& \text { is one of a possible value. }\end{aligned}$

If photons of energy

$12.75eV$ are passing through hydrogen gas in ground state then no of lines in emission spectrum will be

0%
$6$
0%
$4$
0%
$3$
0%
$2$

Explanation

$\begin{array}{l}\text { Given- * Energy of photons }=12.75\mathrm{eV} \\\Rightarrow \text { Let the state of electron initially be } n^{\text {th }} \text { state. } \\\begin{aligned}\Rightarrow \quad 12.75=& 13.6\left[\frac{1}{1}-\frac{1}{n^{2}}\right] \\& \Rightarrow n=4 \\\text { No. of Spectral lines } &=\frac{n(n-1)}{2}=6\end{aligned}\end{array}$

In a hypothetical Bohr hydrogen, the mass of the electron is doubled. The energy

$E_o$ and the radius

$r_o$ of the first orbit will be (

$a_o$ is the bohr radius)

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$E_o=-27.2eV;r_o=a_o/2$
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$E_o=-27.2eV;r_o=a_o$
0%
$E_o=-13.6eV;r_o=a_o/2$
0%
$E_o=-13.6eV;r_o=a_o$

Explanation

$\begin{array}{l}\text { Given-* Mass of the } e^{-}(m) \text { is doubled } \\\text { we know that Energy of an } e^{-} \text {is given by- } \\\qquad E_{0}=-\frac{m e^{4} z^{2}}{8 \varepsilon_{0}^{2} n^{2} h^{2}}=-13.6 \text { ev (at ground state) }\end{array}$

$\begin{array}{l}\text { if mass is doubled then - } \\\qquad E_{0}\text { is also double } \left.\Rightarrow [E_{f}=-27.2\mathrm{eV}\right] \\\text { we also know that radius of the } e^{-} \text {is given by (in an orbit) - } \\\text { orbitat radius re }=\frac{n^{2} h^{2} \varepsilon_{0}}{\pi m z e^{2}}=a_{0}\text { (bohr radius n=1) } \\\left.\qquad[ r=\frac{a_{0}}{2}\right]\text{If the mass is doubled}\end{array}$

The absorption transition between the first and the fourth energy states of hydrogen atom are

$3$ . The emission transition between these states will be:

0%
$3$
0%
$4$
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$5$
0%
$6$

Explanation

Number of emission states are:

$4\,\,to\,\,3$

$4\,\,to\,\,2$

$4\,\,to\,\,1$

$3\,\,to\,\,2$

$3\,\,to\,\,1$

$2\,\,to\,\,1$

So, number of emission states are $6$ .

Two samples of radium, each of mass

$1\ mg$ and

$0.1\ mg$ , are taken to study the emission of alpha particles. It will be observed that

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Both samples will emit the same number of alpha particles
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$1\ mg$ of radium will emit $10$ times the number of alpha particles per second than $0.1\ mg$ does
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Emission of alpha particles is independent of mass of the samples
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$0.1\ mg$ of radium will emit ten times more than that by $$1\ mg of radium

If the difference between

$(n + 1)^{th}$ Bohr radius and

$n^{th}$ Bohr radius is equal to the

$(n -1)^{th}$ Bohr radius then find the value of

$n$

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

Explanation

Radius of

$n^{th} orbit$ is

$r_n=n^2r_0$ where

$r_0$ is radius of ground state.

Given that

$(n+1)^2r_0 - n^2r_0=(n-1)^2r_0$

Solving above equation will result in

$n=4$

Consider an atom of

${Li}^{2+}$ which is a uni-electron system and hence Bohr's model is applicable to it.
If

${r}_{0}=$ radius of first orbit in hydrogen atom, what is the radius of revolution of electron in second excited state in

${Li}^{2+}$ ?

0%
${r}_{0}$
0%
$2{r}_{0}$
0%
$3{r}_{0}$
0%
$9{r}_{0}$

Explanation

$\text{2nd excited state means 3rd normal state i.e n=3}$

$r_n=r_0\dfrac{n^2}{Z}$ ...........(1)

put in the equation-1

$Z=3(Lithium), n=3$ we get

$r_3=3r_0$

Option C is correct.

In an atom an electron excites to the fourth orbit. When it jumps back to the energy levels a spectrum is formed.Total number of spectral lines in this spectrum would be

0%
3
0%
4
0%
5
0%
6

Explanation

In an atom an electron excites to the fourth orbit,

$n=4$

The total number off spectral lines in the spectrum is,

$\dfrac{n(n-1)}{2}=\dfrac{4(4-1)}{2}=\dfrac{4\times 3}{2}=6$

The correct option is D.

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

$5.3\times 10^{-11}$ m. What is the radius of its second orbit?

0%
$21.1\times 10^{-11}$ m.
0%
$31.1\times 10^{-11}$ m.
0%
$51.1\times 10^{-11}$ m.
0%
$81.1\times 10^{-11}$ m.

Explanation

Formula for the radius of orbit is

$r={ r }_{ 0 }\cdot { n }^{ 2 }$

when

$n=2$

$r=5.3\times { 10 }^{ -11 }\times { 2 }^{ 2 }$

$=21.2\times { 10 }^{ -11 }m$

$=2.12\times { 10 }^{ -10 }m$

The wavelengths involved in the spectrum of deuterium

$\left( ^{2}_1D\right)$ are slightly different from the hydrogen spectrum, because:

0%
Sizes of the two nuclei are different
0%
Nuclear forces are different in the two cases
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Masses of the two nuclei are different
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Attraction between the electron and the nucleus in different in the two cases

Explanation

option C mass of

${ _{ 1 }{ H } }^{ 1 }$ and

${ _{ 1 }{ D } }^{ 2 }$ are different hence, the corresponding wavelength are different

A Proton and an

$/alpha$ particle are accelerated through a potential difference of 100V. The ratio of the wavelength associated with the proton to that associated with an

$/alpha$ particle is

0%
$1:2$
0%
$2:1$
0%
$22:1$
0%
$2 \sqrt{2} : 1$

Explanation

$Enrgy=\dfrac{Momentum^2}{2m}=p^2/2m$

So momentum

$p=\sqrt[2]{2mE}$ ,

$m$ is mass.

Wavelength

$\lambda=\dfrac{h}{p} i.e$

$\lambda \propto \dfrac{1}{p}$

So required ratio is

$\sqrt[2]{\dfrac{m_aE_a}{m_pE_p}}$

Mass of alpha particle is

$m_a=4m$ where

$m=m_p$ is mass of proton.

Energy of alpha particle

$E_a=q_aV=2e\times100$

and energy of proton is

$E_p=q_pV=e\times100$

Putting all the values we get the ratio as

$\sqrt[2]{\dfrac{800me}{100me}}=\dfrac{2\sqrt[2]{2}}{1}$

Electrons accelerated from rest by a potential difference of

$12.75V$ , are bombarded on a mono-atomic hydrogen gas. Possible emission of spectral lines are:-

0%
First three Lyman lines, first two Balmer's lines and first Paschen line
0%
First three Lyman lines only
0%
First two Balmer's lines only
0%
None of the above.

Explanation

$\begin{array}{l} \text { when it get energy of } 12.75 \text { ev it excited to } \\ n=4 \quad E_{4}-E_{1}=12.75 \mathrm{ev} \end{array}$

$\begin{array}{l} \text { then after it comes down, possible emission will } \\ \text { be } \end{array}$

$\begin{array}{l} \text { Line fall on } n=1 \text { ane lyman line }=3 \\ \text { Line fall on } n=2 \text { are balmer line }=2 \\ \text { Line fall on } n=3 \text { are paschen line }=1 \end{array}$

2000323_1123433_ans_dcceeebc70954093bc8cdc14caf8ecdb.jpeg

A positroniuim consists of an electron and a positron revolving about their common centre of mass. Calculate the kinetic energy of the electron in ground state:

0%
$1.51\ eV$
0%
$3.4\ eV$
0%
$6.8\ eV$
0%
$13.6\ eV$

Explanation

$E_k \propto m$

Where

$m$ is reduced mass of electron in the orbit.

For hydrogen atom

$m=\dfrac{m_em_p}{m_p +m_e}$ which is

$m_e$ .......(1)

because mass of proton

$m_p$ is much larger

than the mass of electron

$m_e$ .

Mass of positron is equal to mass of electron so reduced mass of electron

for this pair will be

$m=m_e^2/2m_e=m_e/2$ ..............(2)

As energy for hydrogen atom is

$13.6eV$

so energy for electron-positron pair will be

$13.6/2=6.8eV$

Because in this case reduced mass is

$m_e/2$ i.e. half of that in the case of hydrogen atom.(See equation-2 and equation-1)

Option C is correct.

Energy required for the electron excitation in

$Li^{++}$ from the first to the third Bohrs orbit:

0%
$7.625\times { 10 }^{ 28 }mev$
0%
$109 ev$
0%
$8.8\times { 10 }^{ -28 }mev$
0%
$56ev$

Explanation

$\begin{array}{l}\text { we know that energy for exitation of } e^{-}\text {is given as - } \\\qquad E=13.6 \times z^{2}\times\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] \\\text { For exitation from first orbit }\left(n_{1}=1\right) \text { to third orbit }\left(n_{2}=3\right) \\\text { we get }-E=13.6\times(3)^{2}\times\left[\frac{1}{1}-\frac{1}{9}\right]=108.8\mathrm{eV} \\E \approx 109 \mathrm{eV} \\\text {Hence, option (B) is correct. }\end{array}$

The ground state energy of

$H$ - atom is

$-13.6eV$ . The energy needed to ionise

$H$ - atoms from its second excited state is:

0%
$1.51eV$
0%
$3.4eV$
0%
$13.6eV$
0%
$12.1eV$

Explanation

The energy of

$n^{th}$ state in the hydrogen state will be

$E_n=-\dfrac{13.6eV}{n^2}$

so this energy for the second

$excited$ state i.e.

$n=3$ will be

$E_3=\dfrac{-13.6eV}{9}=-1.51ev$

So ionization energy for this state will be

$-(-1.51eV)=1.51eV$

In a sample of hydrogen like atoms all of which are in ground state , a photon beam containing photons of various energies is passed.In absorption spectrum, five dark lies are observed. The number of bright lines in the emission spectrum will be (Assume that all transitions take place)

0%
5
0%
10
0%
15
0%
none of these

A proton, a direction ion and an

$\alpha$ -particle of equal kinetic energy perform circular motion normal to a uniform magnetic field

$B$ . If the radii of their paths are

$r_p, r_d$ and

$r_\alpha$ respectively then [Here,

$q_d = q_p, m_d = 2m_p$ ]

0%
$r_\alpha = r_p > r_d$
0%
$r_\alpha = r_p< r_d$
0%
$r_\alpha > r_d > r_p$
0%
$r_\alpha = r_d = r_p$

Explanation

$\begin{array}{l}\text { Given, } \\\text { a proton } \quad q=+1 \quad m=m_p \\\text { a deutorium } \quad q=+ 1 \quad m=2 \mathrm{m_p} \\\text { a } \alpha \text { particle }\quad q=+ 2 \quad m=4 \mathrm{m_p} \end{array}$

$\begin{array}{l}\text { hane equal Kinctic energy } \\\qquad B=\text { magnetic field. }\end{array}$

$\begin{array}{l}\text { We Know when a change particle enters } \\\text { in a magnetic field it perform } \\\text { circular motion with radius equal to. }\end{array}$

$\begin{aligned}r &=\frac{m v}{q B} \\r &=\frac{\sqrt{2 m K E}}{q B}\end{aligned}$

$\begin{array}{l}r_{\text {proton }}=\frac{\sqrt{2 \times m_p \times K E}}{1\times B} \\\text r_{ deutorium }=\frac{\sqrt{2 \times 2 \times m_p \times K E}}{1 \times B}\\\text r_{\alpha }=\frac{\sqrt{2 \times 4 \times m_p\times K E}}{2 \times B}=\frac{\sqrt{2 \times m_p \times K E}}{1 \times B}\end{array}$

$\text { So, } \quad r_{\text {proton }}=r_{\alpha} < r_{\text {deutorium }}(\text {option } B)$

If radiation of all wavelengths from ultraviolet to infrared is passed through hydrogen gas at room temperature absorption lines will be observed in the

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Lyman series
0%
Balmer's series
0%
Both (1) and (2)
0%
Neither (1) or (2)

Explanation

$\begin{array}{l}\text { For hydrogen atom only one electron is } \\\text { present at ground state. when it absorb } \\\text { energy of different wavelength, it transist } \\\text { from } n=1 \text { to } n=2,3,4 \text { and so on, } \\ \text{so, only lymen series id obsererved due to}\\\text { absence of electron on other energy level. }\end{array}$

The radius of first orbit of hydrogen atom is

$0.53\ A$ . The radius of its fourth orbit will be-

0%
$0.193\ A$
0%
$4.24\ A$
0%
$2.12\ A$
0%
$8.48\ A$

Explanation

$\text{Given- $*$ Radius of first orbit=0.53 A}$

$\text{We know that -}$

$\text{Radius of the orbit of $e^{-}(r),$}$

$\text { r } \alpha n^{2} ,\quad n \rightarrow n^{\text {th }}orbit$

$\Rightarrow \frac{r_{1}}{r_{2}}=\frac{n_{1}^{2}}{n_{2}^{2}}\quad\text { Let } r_{2} \text { be the radius of } 4^{\text {th }} \text { orbit }$

$\Rightarrow \quad r_{2}=\frac{n_{2}^{2}}{n_{1}^{2}} \cdot r_{1}$

$\begin{array}{l}\text { for } 4^{\text {th }} \text { orbit, }n_{2}=4\text { and } n_{1}=1 \\\Rightarrow \quad r_{2}=(4)^{2} \cdot r_{1}=16 \times 0.53 A^{\circ} \\\Rightarrow r_{2}=8.48 A^{\circ}\\\text { Hence, option (D) is correct. }\end{array}$

If in Bohr's atomic model, it is assumed that force between electron the proton varies inversely as

${r^4},$ b energy of the system will be proportional to:

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${n^2}$
0%
${n^4}$
0%
${n^6}$
0%
${n^8}$

Explanation

$\dfrac{mv^2}{r}=\dfrac{k}{r^4}$

$V^2=\dfrac{k}{mr^3}$
we know

$mvr=\dfrac {nh}{2x}$

$m^2v^2r^2=\dfrac{n^2h^2}{4\pi^2}$

$\dfrac{m^2kr^2}{mr^3}=\dfrac{n^2h^2}{4x^2}$

$r=\dfrac{mk4\pi^2}{n^2h^2}$

$k.f$ =

$\dfrac{1}{2}mv^2=\dfrac{1}{2}m/\dfrac{k}{mr^3}$

$=\dfrac{1}{2}m\dfrac{K}{m^1m^3}$

$E\alpha\ n^6$

In a hydrogen atom, the electron revolves round the proton in a circular orbit of radus

$0.528 \mathring {A}$ with a speed of

$2.18 \times 10^6$ m/s. Calculate the centripetal force acting on the electron. (Mass of electron =

$9.1 \times 10^{-31}kg. I \mathring {A}=10^{-10}m)$

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$10.2 \times 10^{-8}$
0%
$8.2 \times 10^{-8}$
0%
$9.2 \times 10^{-8}$
0%
$7.2 \times 10^{-8}$

The wave number of the radiation whose quantum is

$1erg$ is

0%
$5\times {10}^{15}{Cm}^{-1}$
0%
$15\times {10}^{5}{cm}^{-1}$
0%
$1.5\times {10}^{-15}{cm}^{-1}$
0%
$5\times {10}^{-21}{cm}^{-1}$

Explanation

Energy=1erg

$\dfrac {hc}{\lambda}=1erg$

$\dfrac{1}{\lambda}=wave\ number$

$\dfrac{1}{\lambda}=\dfrac{1}{hc}$

$u=\dfrac{1}{hc}$

$u=\dfrac{1}{6.626×10^{−27}×3×10^10}$

$u=\dfrac{100}{19.878}×10^15=5.03×10^{15}cm^{−1}$

The radius of the first orbit of the

$e^-$ of a hydrogen atom is

$5.3\times 10^{11}m$ , what is the radius of its second orbit.?

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$3.12\times 10^{-10}m$
0%
$2.12\times 10^{-10}m$
0%
$4.12\times 10^{-10}m$
0%
$5.12\times 10^{-10}m$

Explanation

Radius

$=\dfrac{x^2}{z}(r_0)$

$n=2,z=1$

$Radius =\dfrac{4}{1}r_0$

$=4\times 5.3\times 10^{-11}m$

$21.2\times 10^{-11}$

$2.12\times 10^{-10}m$

The lifetime of an electron in the state

$n=2$ in the hydrogen atom is about

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$10\ ns$
0%
$1000\ ns$
0%
$1\ ms$
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$10\ ms$

Explanation

The lifetime of an electron in the state

$n=2$ in the hydrogen atom is about

$10ns$ .

The correct option is A.

The moment of momentum of electrons in the third orbit in a hydrogen atom

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$31.67 \times 10^{-34} gm \ m^2/s$
0%
$3.167 \times 10^{-34} kg \ m^2/s$
0%
$31.67 \times 10^{-34} kg \ cm^2/s$
0%
$31.67 \times 10^{-34} gm \ cm^2/s$

CBSE Questions for Class 12 Medical Physics Atoms Quiz 9 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Atoms Quiz 9 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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