The energy of the incident photon is 20 eV and the work function of the photosensitive metal is 10 eV, the stopping potential is

Question

The energy of the incident photon is 20 eV and the work function of the photosensitive metal is 10 eV, the stopping potential is (A) 5 V (B) 10 V (C) 15 V (D) 30 V

Tardigrade Admin · Accepted Answer

V0 = ((20 - 10) eV/e) = (10 × e/e) 10 V

Q. The energy of the incident photon is 20 eV and the work function of the photosensitive metal is 10 eV, the stopping potential is

5 V

10 V

15 V

30 V

$V_{0} = \frac{( 20 - 10 ) e V}{e} = \frac{10 \times e}{e}$ 10 V

Q. The energy of the incident photon is 20 eV and the work function of the photosensitive metal is 10 eV, the stopping potential is

5 V

10 V

15 V

30 V

V0​=e(20−10)eV​=e10×e​ 10 V

$V_{0} = \frac{( 20 - 10 ) e V}{e} = \frac{10 \times e}{e}$ 10 V