The minimum frequency of photon required to break a particle of mass 15.348 amu into $4\alpha$ parti

Question

The minimum frequency of photon required to break a particle of mass 15.348 amu into $4\alpha$ particles is $\_\_\_\_$ kHz .

[mass of He nucleus = 4.002 amu,

$1amu = 1.66 \times 10^{- 27}\text{ }kg,\text{ }h = 6.6 \times 10^{- 34}\text{ }J.s$ and $c = 3 \times 10^{8}\text{ }m/s$ ]

Accepted Answer

$hv = (4 	imes 4.002 - 15.348) 	imes 1.66 	imes 10^{- 27} 	imes \left( 3 	imes 10^{8} ight)^{2}v = 14.94 	imes 10^{19}kHz$

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