The threshold frequency of the metal of the cathode in a photoelectric cell is $1×10^{15} Hz$. 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 $h= 6.63×10^{-34}Js$)
Answer & explanation
Correct answer: option 2
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,
$v=\frac{E}{h}=\frac{8.288×1.6×10^{-19}J}{6.63×10^{-34}Js}=2×10^{15} Hz$