Question

What will be energy in eV of photons of λ - rays having wave length 0.1Å coming out of excited nucleus of radium?

{c = 3 × 10⁸ (m/s), h = 6.625 × 10^–34 (m/s)}

(A) 42.12 × 10⁴

(B) 12.42 × 10⁴

(C) 22.41 × 10⁴

(D) 24.21 × 10⁴

Answer 1

The correct option is (B) 12.42 × 10⁴

Explanation:

λ = 0.1 Å

E = (hc / λ)

E = {(6.625 × 10^–34 × 3 × 10⁸) / (0.1 × 10^–10)}

E = 19.875 × 10^–15 J

as 1J = 1.6 × 10^–19 eV

hence E in eV is {(19.875 × 10^–15) / (1.6 × 10^–19)} = 12.41 × 10⁴ eV

∴ E = 12.41 × 10⁴ eV