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 \(m\) and initial velocity \(v\) collides with a particle \(B\) of mass \(\frac{m}{2}\) which is at rest. The collision is head-on, and elastic. The ratio of the de-Broglie wavelengths \(\lambda_A\) to \(\lambda_B\) after the collision is:
1. \( \frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{1}{3} \)
2. \( \frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=2 \)
3. \( \frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{2}{3} \)
4. \( \frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{1}{2}\)
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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