Consider a beam of electrons (each electron with energy \(E_0\)) incident on a metal surface kept in an evacuated chamber. Then:

1.	no electrons will be emitted as only photons can emit electrons.
2.	electrons can be emitted but all with energy, \(E_0.\)${\mathrm{}}_{}$
3.	electrons can be emitted with any energy, with a maximum of \({E}_0-\phi\) (\(\phi\) is the work function).
4.	electrons can be emitted with any energy, with a maximum \(E_0.\)

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:

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 work functions for metals \(A,B,\) and \(C\) are respectively \(1.92\) eV, \(2.0\) eV, and \(5\) eV. According to Einstein's equation, the metals that will emit photoelectrons for a radiation of wavelength \(4100~\mathring{A}\) is/are:

An electron is accelerated from rest through a potential difference of \(V\) volt. If the de Broglie wavelength of an electron is $\mathrm{}$\(1.227\times10^{-2}~\text{nm}.\) What will be its potential difference?

In an experiment of the photoelectric effect, the wavelength of incident radiation is $\lambda$. The wavelength of incident radiation is reduced to $\frac{1}{3}$rd of initial value and the maximum kinetic energy of photoelectron is observed to be n times the previous value. What will be the threshold wavelength for the metal plate?