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#### The graph which depicts the results of Rutherford gold foil experiment with $\alpha -$particles is:  $\theta$: Scattering angle Y : Number of scattered $\alpha -$ particles detected (Plots are schematic and not to scale)   Option: 1 Option: 2  Option: 3  Option: 4

As $\dpi{120} Y= N(\theta) \propto \frac{1}{\sin ^{4}\left(\frac{\theta}{2}\right)}$

Here when $\theta=0$ Sin will be zero then Y will tend to infinity.

and that is satisfied by the only option 1

So the correct option is 1.

#### The time period of revolution of electron in its ground state orbit in a hydrogen atom is $1.6\times10^{-16}s$. The frequency of revolution of the electron in its first excited state (in s-1) is:- Option: 1  Option: 2  Option: 3    Option: 4

The velocity of the nth orbit

$V_n \ \alpha \ \frac{z}{n}$

the radius of the nth orbit

$r_n \ \alpha \ \frac{n^2}{z}$

$T=\frac{2\pi r}{V}\\ \\ \Rightarrow T \ \propto \ \frac{n^2\times n}{z \times z}=\frac{n^3}{z^2}\\\\ \Rightarrow T \propto\frac{1}{f} \\ \Rightarrow f \ \propto \ \frac{z^2}{n^3}$

$f_1=\frac{1}{1.6\times 10^{-16}} \ s^{-1}$

and

$\frac{f_1}{f_2}=(\frac{n_2}{n_1})^{3}=8\\ \Rightarrow f_2=\frac{f_1}{8}=7.8 \times 10^{14} \ s^{-1}$

So, option (3) is correct.

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#### The energy required to ionised a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength (in nm) of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state ? Option: 1 11.4 Option: 2 24.2 Option: 3 8.6 Option: 4 35.8

$\frac{hc}{\lambda }=(13.6eV)z^{2}\left [ \frac{1}{n^{2}_{1}}-\frac{1}{n^{2}_{2}} \right ]$

$n_{1}=1$

$n_{2}=3$

$\frac{hc}{\lambda }=(13.6eV)(3^{2})\left [ \frac{1}{1^{2}}-\frac{1}{3^{2}} \right ]$

$\Rightarrow \frac{hc}{\lambda }=(13.6eV)(9)\left ( \frac{8}{9} \right )$

Wavelength$=\frac{1240}{8\times 13.6}nm$

${\lambda }=11.40nm$

Hence the correct option is (1).

#### The first member of the Balmer series of the hydrogen atom has a wavelength of 6561 $\AA$ . The wavelength of the second member of the Balmer series (in nm ) is _______ . Option: 1 486 Option: 2 972 Option: 3 986 Option: 4 586

\begin{aligned} \frac{1}{\lambda} &=\mathrm{R} Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \\ \frac{1}{\lambda_{1}} &=R(1)^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5 \mathrm{R}}{36} \\ \frac{1}{\lambda_{2}} &=\mathrm{R}(1)^{2}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3 \mathrm{R}}{16} \\ \frac{\lambda_{2}}{\lambda_{1}} &=\frac{20}{27} \\ \lambda_{2} &=\frac{20}{27} \times 6561 \mathrm{A}=4860 \mathrm{A} \\ &=486 \mathrm{nm} \end{aligned}

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#### The activity of a radioactive sample falls from  to   in 30 minutes. Its half life ( in minutes ) is close to :   Option: 1 62 Option: 2 72 Option: 3 66 Option: 4 52

Activity at any time is given as

$A=A_0e^{-\lambda t}\\ \Rightarrow \frac{A}{A_0}=e^{-\lambda t}\\ \Rightarrow ln( \frac{A_0}{A})=\lambda t$

After half-life,

$\\t=T_{\frac{1}{2}} \\ A=\frac{A_0}{2}$

so $\ln 2=\lambda T_{1 / 2}$

Now for $t=30 \ min \ , A_0=700 \ s^{-1}, \ A=500 \ s^{-1}$

$\ln \left[\frac{700}{500}\right]=\lambda(30 )$

from( (i) and (ii)

$\frac{\ln 2}{\ln (7 / 5)}=\frac{T_{1 / 2}}{(30 )}$

${t_{1 / 2}} =61.81 \cong 62 \ min$

Hence the correct option is (1).

#### A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance 200  is first charged and later connected with resistor If the ratio of charge on capacitor to the activity of radioactive sample is fixed with respect to time then the value of 'R' should be ______.

$\begin{gathered} t_{\text {mean }}=30 \times 10^{-3} \mathrm{~S}=\frac{1}{\lambda} \\ C=200 \mu \mathrm{F} \end{gathered}$

Let $q_{0}$ be the initial change on the capacitor

$\rightarrow q= q_{0}e^{-t/\tau }$

Discharging of capacitor condition.

Activity after time  t

$A_{t}=A_{0} e^{-\lambda t}$

\begin{aligned} &\frac{q_{t}}{A_{t}}=\frac{q_{0}}{A_{0}} \frac{e^{-t / \tau}}{e^{-\lambda t}} \\ &\frac{q_{t}}{A_{t}}=\frac{q_{0}}{A_{0}} e^{(-t / \tau+\lambda t)}=K \end{aligned}

\begin{aligned} \therefore-\frac{t}{\tau} &+\lambda t=0 \\ \frac{1}{\tau} &=\lambda \\ \frac{1}{R C} &=\lambda \end{aligned}

\begin{aligned} t_{\text {mean }}=\frac{1}{\lambda} &=R C \\ \end{aligned}

\begin{aligned} 30 \times 10^{-3}=R \times 200 \times 10^{-6} \\ \end{aligned}

\begin{aligned} R &=\frac{30}{200} \times 10^{+3} \\ \\R &=150 \Omega \end{aligned}

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#### An electron and proton are separated by a large distance. The electron starts approaching the proton with energy The proton captures the electrons and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength . What is the maximum kinetic energy of the emitted photoelectron?   Option: 1   Option: 2 Option: 3 Option: 4

Energy lost by electron=Energy of emitted photon

\begin{aligned} &\left(E_{i}-E_{f}\right)=\frac{h c}{\lambda} \\ &\left(3 \operatorname{ev}-\left(\frac{-13.6 \mathrm{eV}}{3^{2}}\right)\right]=\frac{h c}{\lambda} \\ \end{aligned}

\begin{aligned} &{\because \text { for second excitcd state }} \\ &\text { i.e., } For \ (n=3) \Rightarrow E_{n}=\frac{-13.6 \mathrm{eV}}{n^{2}}=\frac{-13.6 \mathrm{eV}}{3^{2}}=-1.51 eV \\ &\frac{h c}{\lambda}=4.51 \mathrm{eV} \end{aligned}

By Einstein's photoelectric equation
\begin{aligned} &\frac{h c}{\lambda}=\phi_{0}+K E_{\max } \\ &4.51 \mathrm{ev}=3.1 \mathrm{ev}+K E_{\text {max }} \\ &K E_{\max }=1.41 \mathrm{eV} \end{aligned}
The correct option is (2)

#### Consider the following statements : A. Atoms of each element emit characteristics spectrum. B. According to Bohr's Postulate, an electron in a hydrogen atom, revolves in a certain stationary orbit. C. The density of nuclear matter depends on the size of the nucleus. D. A free neutron is stable but a free proton decay is possible. E. Radioactivity is an indication of the instability of nuclei. Choose the correct answer from the options given below : Option: 1 A, B, C, D and E Option: 2 A, B and E only Option: 3 B and D only Option: 4 A, C and E only

$\rightarrow$ A free neutron is unstable
$\rightarrow$ Density of nuclear matter is constant for all elements.
The correct option is (2)

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#### The   of molybdenum has wavelength . If the energy of a molybdenum atoms with a electron knocked out is , the energy of this atom when an L electron is knocked out will be ________ (Round off to the nearest integer)

$E_{1}$ - Energy when an L shell-electron is knocked out
$E_{2}$ - Energy when a K shell-electron is knocked out
$E_{2}=27.5 \mathrm{keV}$
$K_{\alpha} \rightarrow$wavelength emitted during knocking out of L-electron
$E_{K_{\alpha}} =\frac{1240}{\lambda_{k\alpha}}=\frac{1240\left ( ev \right )}{0.071}$
$E_{k \alpha }=\frac{1240}{71 \times 10^{-3}}=\frac{1240 \times 10^{3}}{71}(\mathrm{ev})$

$E_{2}=E_{1}+E_{k_{\alpha}}$

$27.5 \mathrm{eV}=E_{1}+17.5 \mathrm{keV}$

$E_{1}=10eV$

#### A particular hydrogen-like ion emits radiation of frequency when it makes the transition from n = 3 to n = 1, The frequency in Hz of radiation emitted in transition from n =2 to n= 1 will be :Option: 1Option: 2Option: 3Option: 4

$f= 2\cdot 92\times 10^{15}Hz$
$\Delta E= hf= Rchz^{2}\left [ \frac{1}{n_f^{2}}-\frac{1}{n_i^{2}} \right ]$
$hf_{1}= Rchz^{2}\left [ \frac{1}{1^{2}}-\frac{1}{3^{2}} \right ]$
$hf_{2}= Rchz^{2}\left [ \frac{1}{1^{2}}-\frac{1}{2^{2}} \right ]$
$\frac{f_{1}}{f_{2}}= \frac{\frac{8}{9}}{\frac{3}{4}}= \frac{32}{27}$
${f_{2}}= \left ( \frac{27}{32} \right )\times 2\cdot 92\times 10^{15}$
${f_{2}}=2\cdot 46\times 10^{15}$
The correct option is (2)