The threshold frequency of the metal of the cathode in a photoelectric cell is $1×10^{16} Hz$. When a certain beam of light is incident on the cathode, it is found that a stopping potential 4.144 V is required to reduce the current to zero. The frequency of the incident radiation is: (Take $h= 6.63×10^{-34}Js$)

$2×10^{15} Hz$

Here,

Threshold frequency, $v_0=1×10^{15} Hz$

Stopping potential, $V_s=4.144V$

Work function, $\phi_0=hv_0$

$=\frac{6.63×10^{-34}×1×10^{15}}{1.6×10^{-19}}eV=4.144eV$

According to Einstein’s photoelectric equation

$hv=\phi_0+eV_s$

= 4.144 eV + 4.144 eV = 8.288 eV

Energy of incident photon, E = hv = 8.288eV

∴ Frequency of incident photon,

$v=\frac{E}{h}=\frac{8.288×1.6×10^{-19}J}{6.63×10^{-34}Js}=2×10^{15} Hz$