If V be the accelerating voltage, then the maximum frequency of continuous x-rays is given by

1.  $\frac{eh}{V}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$

2.  $\frac{hV}{e}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$

3.  $\frac{eV}{h}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$

4. $\frac{h}{eV}$

When a metallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2$\lambda$, the stopping potential is $\frac{V}{4}$ .The threshold wavelength for metallic surface is:

1. 5$\lambda$               
2. $\frac{5}{2}$$\lambda$
3. 3$\lambda$               
4. 4$\lambda$

Photoelectric emission occurs only when the incident light has more than a certain minimum 

1. wavelength

2. intensity

3. frequency

4. power

The threshold frequency for a photo-sensitive metal is $3.3\times {10}^{14}<mspace\; width="1em"></mspace>\mathrm{Hz}.$ If the light of frequency $8.2\times {10}^{14}<mspace\; width="1em"></mspace>\mathrm{Hz}$ is incident on this metal, the cut-off voltage for the photo-electric emission is nearly:

1. 2 V                                        2. 3 V

3. 5 V                                         4. 1 V