The correct graph respectively the relation between energy (E) of photoelectrons and frequency v of incident light is

1.  

2.  

3.  

4.  

 

The stopping potential applied between a photocathode and respective anode is such that the fastest electron can fly only one half of the distance L between cathode and anode. For the same stopping potential, the distance between cathode and anode is reduced to L/2. The fastest electron can:

1.  reach the anode

2.  fly a distance greater than L/4

3.  fly a distance less than L/4

4.  fly a distance L/4

If the stationary proton and $\mathrm{\alpha }$-particle are accelerated through same potential difference, the ratio of de Broglie's wavelength will be

1.  $2$:1

2.  1:1

3.  $2\sqrt{2}$:1

4.  none of these

Radiation of energy \(E\) falls normally on a perfectly reflecting surface. The momentum transferred to the surface is:
(\(c\) = velocity of light)

The work functions for metals A, B and C are respectively 1.92 eV, 2.0 eV and 5eV. According to Einstein's equation, the metals which will emit photo electrons for a radiation of wavelength 4100$\stackrel{\circ }{\mathrm{A}}$ is/are-

1. None

2. A only

3. A and B only

4. All the three metals

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. \(1~\mathring{A}\)
2. \(0.1~\mathring{A}\)
3. \(10~\mathring{A}\)
4. \(0.01~\mathring{A}\)

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