Q. No.The radiation corresponding to \(3\rightarrow 2\) transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of \(3\times 10^{-4}~\text{T}\). If the radius of the largest circular path followed by these electrons is \(10.0~\text{mm}\), the work function of the metal is close to:
1. \(1.1~\text{eV}\)
2. \(0.8~\text{eV}\)
3. \(1.6~\text{eV}\)
4. \(1.8~\text{eV}\)
1. \(1.2~\mathring{A}\)
2. \(2.4~\mathring{A}\)
3. \(0.5~\mathring{A}\)
4. \(1.7~\mathring{A}\)
| 1. | \(\mathrm{1\over 4}\)th of the de-Broglie wavelength of the electron in the ground state. |
| 2. | four times the de-Broglie wavelength of the electron in the ground state |
| 3. | two times the de-Broglie wavelength of the electron in the ground state |
| 4. | half of the de-Broglie wavelength of the electron in the ground state |
Radiation with wavelength \(\lambda\) is incident on a photocell, causing the fastest emitted electron to have a speed \(v.\) If the wavelength is changed to \(\dfrac{3\lambda}{4},\) what will be the speed of the fastest emitted electron?
1. \( >v\left(\dfrac{4}{3}\right)^{\frac{1}{2}} \)
2. \( <v\left(\dfrac{4}{3}\right)^{\frac{1}{2}} \)
3. \( =v\left(\dfrac{4}{3}\right)^{\frac{1}{2}} \)
4. \( =v\left(\dfrac{3}{4}\right)^{\frac{1}{2}}\)
An electron beam is accelerated by a potential difference \(V\) to hit a metallic target to produce \({X}\)-ray. It produces continuous as well as characteristic \({X}\)-rays. If \(\lambda_{\text{min}}\) is the smallest possible wavelength of \({X}\)-ray in the spectrum, the variation of \(\log \lambda_{\text{min}}\) with \(\log V\) is correctly represented in:
| 1. |  |
2. |  |
| 3. |  |
4. |  |
A particle \(A\) of mass \(m\) moving with speed \(v\) collides elastically and head-on with another particle \(B\) of mass \(\dfrac{m}{2},\) which is initially at rest. After the collision, the de-Broglie wavelengths of particles \(A\) and \(B\) are \(\lambda_A\) and \(\lambda_B\) respectively. What is the value of the ratio \(\dfrac{\lambda_A}{\lambda_B} \text{?}\)
| 1. | \( \dfrac{\lambda_{{A}}}{\lambda_{{B}}}=\dfrac{1}{3} \) | 2. | \( \dfrac{\lambda_{{A}}}{\lambda_{{B}}}=2 \) |
| 3. | \( \dfrac{\lambda_{{A}}}{\lambda_{{B}}}=\dfrac{2}{3} \) | 4. | \( \dfrac{\lambda_{{A}}}{\lambda_{{B}}}=\dfrac{1}{2}\) |
When monochromatic light of frequency \(n\) is incident on a metal surface, the maximum speed of the emitted photoelectrons is \(v.\) If the frequency of the incident light is increased to \(3n,\) what will be the new maximum speed of the emitted photoelectrons?
1. more than \(\sqrt3{v}\)
2. equal to \(\sqrt3{v}\)
3. less than \(\sqrt3{v}\)
4. \({v}\)
[Take Planck's constant \({h}=6.62\times10^{-34}~\text{Js}]\)
1. \(10^{19}\)
2. \(10^{22}\)
3. \(10^{18}\)
4. \(10^{20}\)
1. \(\frac{1}{\lambda_{com}}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2}\)
2. \(\lambda_{com}=\frac { 2 \lambda_1 \lambda_2 }{ \sqrt {\lambda^2_1 + \lambda^2_2}}\)
3. \(\lambda_{com}= \lambda_1= \lambda_2\)
4. \(\lambda_{com} = \bigg(\frac{\lambda_1 +\lambda_2}{2}\bigg)\)
Two particles move at the right angle to each other. Their de-Broglie wavelengths are \(\lambda_1\) and \(\lambda_2\) respectively. The particles suffer perfectly inelastic collision. The de-Broglie wavelength \(\lambda\), of the final particle, is given by:
1. \( \lambda=\frac{\lambda_1+\lambda_2}{2} \)
2. \( \frac{1}{\lambda^2}=\frac{1}{\lambda_1^2}+\frac{1}{\lambda_2^2} \)
3. \( \frac{2}{\lambda}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2} \)
4. \( \lambda=\sqrt{\lambda_1 \lambda_2}\)
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