The minimum frequency of photon required to break a particle of mass 15.348 amu into 4 particles is _______ kHz. [Mass of He nucleus = 4.002 amu, 1 amu = kg, J.s and m/s] [2026]

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Solution

Q.
The minimum frequency of photon required to break a particle of mass 15.348 amu into 4 $α$ particles is _______ kHz.

[Mass of He nucleus = 4.002 amu, 1 amu = $1.66 \times 10^{- 27}$ kg, $h = 6.6 \times 10^{- 34}$ J.s and $c = 3 \times 10^{8}$ m/s] [2026]

1 $14.94 \times 10^{19}$

2 $9 \times 10^{19}$

3 $9 \times 10^{20}$

4 $14.94 \times 10^{20}$

Ans.
(1)

$h ν = (4 \times 4.002 - 15.348) \times 1.66 \times 10^{- 27} \times {(3 \times 10^{8})}^{2}$

$ν = 14.94 \times 10^{19} kHz$

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