JEE Mains · Physics · STD 12 - 13. Nuclei
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,
\(1 amu =1.66 \times 10^{-27} kg, h =6.6 \times 10^{-34} J . s\) and \(c =\) \(\left.3 \times 10^8 m / s \right]\)
- A \(9 \times 10^{19}\)
- B \(9 \times 10^{20}\)
- C \(14.94 \times 10^{20}\)
- D \(14.94 \times 10^{19}\)
Answer & Solution
Correct Answer
(D) \(14.94 \times 10^{19}\)
Step-by-step Solution
Detailed explanation
\( hv=(4\times4.002-15.348)\times1.66\times10^{-27}\times(3\times10^{8})^{2} \) \( v=14.94\times10^{19}kHz \)
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