A photoelectric surface is illuminated successively by the monochromatic light of wavelength \(\lambda\) and \(\frac{\lambda}{2}\). If the maximum kinetic energy of the emitted photoelectrons in the second case is \(3\) times that in the first case, the work function of the surface of the mineral is:
[\(h\) = Plank’s constant, \(c\) = speed of light]
1. \(\frac{hc}{2\lambda}\)
2. \(\frac{hc}{\lambda}\)
3. \(\frac{2hc}{\lambda}\)
4. \(\frac{hc}{3\lambda}\)

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. \(1.5~\text{eV}\)

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)

The threshold frequency for a photosensitive metal is \(3.3\times10^{14}~\text{Hz}\). If the light of frequency \(8.2\times10^{14}~\text{Hz}\) is incident on this metal, the cutoff voltage for the photoelectric emission will be:

The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be:

1. 2.4 V

2. -1.2 V

3. -2.4 V

4. 1.2 V

When photons of energy h$\nu$ fall on an aluminium plate (of work function E₀), photoelectrons of maximum kinetic energy K are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be:

1. $K+E_0$

2. 2K

3. K

4. K+h$\nu$