The work function of cesium is 2.27 eV. The cut off voltage which stops the emission of electron from cesium when irradiated with light of wavelength 400 nm is

0.84 V

The correct answer is Option (1) → 0.84 V

To find cut-off potential, one can use photoelectric equation -

$K_{max}=E_{photon}-\phi$   ...(1)

where,

$K_{max}$, Maximum kinetic energy of the emitted electron

$E_{photon}$, Energy of incident photon

$\phi$, Work function of cesium = $2.27×1.6×10^{-19}J=3.635×10^{-19}J$

$λ$, Wavelength of incident light = 400 nm

$E_{photon}=\frac{hc}{λ}=\frac{6.626×10^{-34}×3×10^8}{400×10^{-9}}$

$=4.97×10^{-19}J$

$K_{max}=E_{photon}-\phi$

$=4.97×10^{-19}J-3.635×10^{-19}J$

$=1.335×10^{-19}J$

Now,

$K_{max}=eV_{cutoff}$

$V_{cutoff}=\frac{K_{max}}{e}=\frac{1.335×10^{-19}}{1.602×10^{-19}}=0.64V$