Question

The work function of a metallic surface is 5.01 eV. The photoelectrons are emitted when light of wavelength 2000 Å falls on it. The potential difference applied to stop the fastest photoelectrons is - [h = 4.14 × 10^–15eVs]

• A. 1.2 V
• B. 2.24 V
• C. 3.6 V
• D. 4.8 V

Answer 1

Answer: (A)

1.2 V

Answer 2

eV₀ = hn – f

eV₀ = $\left( \frac { 12375 } { 2000 } - 5.01 \right)$eV

V₀ = 1.2 V