A certain metallic surface is illuminated with monochromatic light of wavelength \(\lambda\). The stopping potential for photoelectric current for this light is \(3V_0\). If the same surface is illuminated with a light of wavelength \(2\lambda\)$\mathrm{}$, the stopping potential is \(V_0\). The threshold wavelength for this surface for the photoelectric effect is:

1.	\(6\lambda\)	2.	\(4\lambda\)
3.	\(\dfrac{\lambda}{4}\)	4.	\(\dfrac{\lambda}{6}\)

Which of the following figures represent the variation of the particle momentum and the associated de-Broglie wavelength?

1.		2.
3.		4.

When the energy of the incident radiation is increased by \(20\%\), the kinetic energy of the photoelectrons emitted from a metal surface increases from \(0.5~\text{eV}\) to \(0.8~\text{eV}\). The work function of the metal is:
1. \(0.65~\text{eV}\)
2. \(1.0~\text{eV}\)
3. \(1.3~\text{eV}\)
4. \(1.5~\text{eV}\)

If the kinetic energy of the particle is increased to \(16\) times its previous value, the percentage change in the de-Broglie wavelength of the particle is:
1. \(25\)
2. \(75\)
3. \(60\)
4. \(50\)

For photoelectric emission from certain metals, the cutoff frequency is \(\nu\). If radiation of frequency \(2\nu\)$\mathrm{}$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be:
(\(m\) is the electron mass)

A \(200~\text{W}\) sodium street lamp emits yellow light of wavelength \(0.6~\mu\text{m}\)$$. Assuming it to be \(25\%\) efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is:
1. \(1.5\times 10^{20}\)
2. \(6\times 10^{18}\)
3. \(62\times 10^{20}\)
4. \(3\times 10^{19}\)

An \(\alpha\text-\)particle moves in a circular path of radius \(0.83~\text{cm}\) in the presence of a magnetic field of \(0.25~\text{Wb/m}^2\). The de-Broglie wavelength associated with the particle will be:
1. \(1~\mathring{\text{A}}\)
2. \(0.1~\mathring{\text{A}}\)
3. \(10~\mathring{\text{A}}\)
4. \(0.01~\mathring{\text{A}}\)

A radioactive nucleus of mass M emits a photon of frequency $\nu$ and the nucleus will recoil. The recoil energy will be:

1.  $\frac{{\mathrm{h}}^2{\mathrm{\nu }}^2}{2{\mathrm{Mc}}^2}$

2.  zero

3.  $\frac{\mathrm{h\nu }}{\mathrm{c}\sqrt{2\mathrm{M}}}$

4.  $\frac{\mathrm{c}\sqrt{2\mathrm{M}}}{\mathrm{h\nu }}$