Question

47. The energy of an a-particle, whose de-Broglie wavelength is 0.004 Å is (A) 1270 eV (B) 1270 keV (C) 1270 MeV (D) 1270 GeV​

Answer 1

Answer:

The correct option is (A) 1280 eV

Explanation:

λ = 0.004 Å = 0.004 × 10–10 m = 4 × 10–13 m.

Given is α particle, m = 6.64 × 10–27 kg

We know λ = (h/P) = {h / √(2mE)}

∴ √(2mE) = (h/λ)

∴ E = (h2 / λ2) × (1 / 2m)

∴ Energy = E = [(6.62 × 10–34)2 / (4 × 10–13)2] × [1 / {2 × 6.64 × 10–27 kg}]

= 0.206 × 10–15

= 2060 × 10–19 J

we know 1J = 1.6 × 10–19 eV

∴ E in eV is 1287.5 eV ≈ 1280 eVRead