A light of wavelength $\lambda$ is incident on the metal surface and the ejected fastest electron has speed $v.$ If the wavelength is changed to $\frac{3\lambda}{4},$ then the speed of the fastest emitted electron will be:

The current conduction in a discharge tube is due to:
1.  electrons only
2.  +ve ions and –ve ions
3.  –ve ions and electrons
4.  +ve ions and electrons

If a light of amplitude A and wavelength λ is incident on a metallic surface, then the saturation current flow is proportional to (assume cut-off wavelength = ${\lambda }_0$):

1. $A^2,$ $if$ $\lambda$ $>$ ${\lambda }_0$

2. $A^2,$ $if$ $\lambda$ $<$ ${\lambda }_0$

3. $A,$ $if$ $\lambda$ $>$ ${\lambda }_0$

4. $A,$ $if$ $\lambda$ $<$ ${\lambda }_0$

The total energy of an electron is $3.555~\text{MeV}.$ Its kinetic energy will be:
1.  $3.545~\text{MeV}$
2.  $3.045~\text{MeV}$
3.  $3.5~\text{MeV}$
4.  none of the above

What did Einstein prove by the photo-electric effect?
1. $E = h\nu$
2. $K.E = \frac{1}{2}mv^2$
3. $E= mc^2$
4. $E = \frac{-Rhc^2}{n^2}$

A photo-cell is illuminated by a source of light, which is placed at a distance $d$ from the cell. If the distance becomes $\dfrac{d}{2}$, then the number of electrons emitted per second will be: 
1. same
2. four times
3. two times
4. one-fourth