Circular MotionHard
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 $c = 3 \times 10^{8}\text{ }m/s$ ]
Options
A.$9 \times 10^{19}$
B.$9 \times 10^{20}$
C.$14.94 \times 10^{20}$
D.$14.94 \times 10^{19}$
Solution
$hv = (4 \times 4.002 - 15.348) \times 1.66 \times 10^{- 27} \times \left( 3 \times 10^{8} \right)^{2}v = 14.94 \times 10^{19}kHz$
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