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. \(1.8~\text{eV}\)

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. \( \frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{1}{2}\)

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. \( \lambda=\sqrt{\lambda_1 \lambda_2}\)