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 = $4.002 \mathrm{amu}, 1 \mathrm{amu}=1.66 \times 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J}. \mathrm{s}$ and $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ]

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 \mathrm{amu}, 1 \mathrm{amu}=1.66 \times 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J}. \mathrm{s}

and

\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}

]

Quantrex Academy · Accepted Answer

Correct answer: A. 14.94 × 10^ 19

h v =(4 × 4.002-15.348) × 1.66 × 10^ -27 ×(3 × 10^8)^2 v =14.94 × 10^ 19 kHz

The minimum frequency of photon required to break a particle of...