Question

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 Å is 0.42 V. If the threshold frequency is x × 10¹³/s, where x is ____ (nearest integer).
(Given, speed of light = 3 × 10⁸ m/s, Planck’s constant = 6.63 × 10^–34 Js)

Answer 1

Correct Answer: 35

The correct answer is 35
\(\frac{hc}{λ} - Ø = KE = eV_0 \)
\(⇒ \frac{6.63 × 10^{-34} × 3 × 10^8}{6630 × 10^{-10}} - 6.63 × 10^{-34}\)
\(=  ƒ_{th} = 1.6 × 10^{-19} × 0.4\)
\(⇒ ƒ_{th} ≃35.11 × 10^{13} H\)