The minimum frequency of photon required to break a particle of mass 15.348 amu into 4α particles i $s ____$ kHz.
[mass of He nucleus = 4.002 amu, 1 amu = 1.66 × 10⁻²⁷ kg, h = 6.6 × 10⁻³⁴ J.s and c = 3 × 10⁸ m/s]

Question

The minimum frequency of photon required to break a particle of mass 15.348 amu into 4α particles i $s ____$ kHz.

[mass of He nucleus = 4.002 amu, 1 amu = 1.66 × 10⁻²⁷ kg, h = 6.6 × 10⁻³⁴ J.s and c = 3 × 10⁸ m/s]

G1 Physics · Accepted Answer

The initial particle mass is M(initial) = 15.348 amu. The total rest mass of four α-particles is M(final) = 4 × 4.002 = 16.008 amu. Therefore the required mass increase is Δm = 16.008 - 15.348 = 0.660 amu. The minimum photon energy must supply this rest-energy difference: hν = Δmc². Thus, ν = Δmc²/h. Substitute: ν = [0.660 × 1.66 × 10⁻²⁷ × (3 × 10⁸)²]/(6.6 × 10⁻³⁴) ≈ 1.494 × 10²³ Hz. Convert Hz to kHz: ν = 1.494 × 10²⁰ kHz = 14.94 × 10¹⁹ kHz. Correct option: D.

The minimum frequency of photon required to break a particle of mass 15.348 amu into 4α particles i s ____ kHz. [mass of He nucleus = 4.002 amu, 1 amu = 1.66 × 10⁻²⁷ kg, h = 6.6 × 10⁻³⁴ J.s and c = 3 × 10⁸ m/s]