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

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.

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Posted by

vishal kumar

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 $5.6\times 10^{12}$
Option: 2 $1.6\times 10^{14}$
Option: 3 $7.8\times 10^{14}$
Option: 4 $6.2\times 10^{15}$

The velocity of the n_th orbit

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

the radius of the n_th 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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Ritika Jonwal

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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).

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avinash.dongre

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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vishal kumar

JEE Main high-scoring chapters and topics

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The activity of a radioactive sample falls from $700 \mathrm{s}^{-1}$ to $500 \mathrm{s}^{-1}$ 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).

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Ritika Jonwal

A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance 200 $\mu F$ is first charged and later connected with resistor $'R'.$ 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 ______ $\Omega$ .

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

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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vishal kumar

NEET 2024 Most scoring concepts

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An electron and proton are separated by a large distance. The electron starts approaching the proton with energy $3 \mathrm{eV}.$ 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 $4000 \AA$ . What is the maximum kinetic energy of the emitted photoelectron?

Option: 1 $7.61 \mathrm{eV}$

Option: 2 $1.41 \mathrm{eV}$
Option: 3 $3.3 \mathrm{eV}$
Option: 4 $\text { No photoelectron would be emitted }$

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)

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vishal kumar

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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vishal kumar

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The $\mathrm{K}_{\alpha} \mathrm{X}-ray$ of molybdenum has wavelength $0.071 \mathrm{~nm}$ . If the energy of a molybdenum atoms with a $\mathrm{K}$ electron knocked out is $27.5 \mathrm{keV}$ , the energy of this atom when an L electron is knocked out will be ________ $\mathrm{keV}.$ (Round off to the nearest integer)
$\left[\mathrm{h}=4.14 \times 10^{-15} \mathrm{eVs}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right]$

$E_{1}$ - 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
$E_{K_{\alpha}} =\frac{1240}{\lambda_{k\alpha}}=\frac{1240\left ( ev \right )}{0.071}$
$E_{k \alpha }=\frac{1240}{71 \times 10^{-3}}=\frac{1240 \times 10^{3}}{71}(\mathrm{ev})$

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

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

$E_{1}=10eV$

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vishal kumar

A particular hydrogen-like ion emits radiation of frequency $2. 92\times 10^{15}Hz$ 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: 1 $6\cdot 57\times 10^{15}$
Option: 2 $2\cdot 46\times10^{15}$
Option: 3 $0\cdot 44\times10^{15}$
Option: 4 $4\cdot 38\times10^{15}$

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

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Get Answers to all your Questions

All Questions

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

Posted by

vishal kumar

The time period of revolution of electron in its ground state orbit in a hydrogen atom is . The frequency of revolution of the electron in its first excited state (in s-1) is:- Option: 1 Option: 2 Option: 3 Option: 4

Posted by

Ritika Jonwal

Crack CUET with india's "Best Teachers"

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

Posted by

avinash.dongre

The first member of the Balmer series of the hydrogen atom has a wavelength of 6561 . 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

Posted by

vishal kumar

JEE Main high-scoring chapters and topics

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

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

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

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

The activity of a radioactive sample falls from $700 \mathrm{s}^{-1}$ to $500 \mathrm{s}^{-1}$ in 30 minutes. Its half life ( in minutes ) is close to :

Option: 1 62
Option: 2 72
Option: 3 66
Option: 4 52