Question

The difference of kinetic energy of photoelectrons emitted from a surface wavelength 2500 Å and 5000 Å will be 

(A) 1.98 × 10^–19 J 

(B) 1.98 × 10^–19 J 

(C) 3.96 × 10^–19 eV 

(D) 3.96 × 10^–19 J.

Answer 1

The correct option is (D) 3.96 × 10^–19 J.

Explanation:

Kinetic energy = Ek = (hc / λ)

E_k1 – E_k2 = hc {(1 / λ₁) – (1/λ₂)}

∴ E_k1 – E_k2 = 6.62 × 10^–34 × 3 × 10⁸ [{1 / (2500 × 10^–10)} – {1 / (5000 × 10^–10)}]

= 19.86 × 10^–26 × 10¹⁰ [{5000 – 2500} / {1250 × 10⁴}]

= 39.72 × 10^–20 J

= 3.97 × 10^–19 J