Question: Q. 1. The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with $1 \mathrm{MeV}$ energy is nearly (a) $1.2 \mathrm{~nm}$ (b) $1.2 \times 10^{-3} \mathrm{~nm}$ (c) $1.2 \times 10^{-6} \mathrm{~nm}$ (d) $1.2 \times 10 \mathrm{~nm}$
[NCERT Exemplar]
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Solution:
Ans. Correct option : (b)
Explanation: Energy of the photon must be equal to the binding energy of proton
So, energy of photon $=1 \mathrm{MeV}=10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}$
$$ \begin{aligned} \lambda & =\frac{h c}{E}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-13}}=\frac{6.63 \times 3}{1.60} \times 10^{-26+13} \ & =\frac{19.89}{1.60} \times 10^{-13}=12.4 \times 10^{-13}=1.24 \times 10^{1} \times 10^{-13} \ & =1.24 \times 10^{-9} \times 10^{-3}=1.24 \times 10^{-3} \mathrm{~nm} \end{aligned} $$
