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The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly:
A. $1.2nm$
B. $1.2 \times {10^{ - 3}}nm$
C. $1.2 \times {10^{ - 6}}nm$
D. $1.2 \times {10^1}nm$

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Last updated date: 21st Apr 2024
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Answer
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Hint: Equate the energy given to remove the proton from its nucleus with the energy of the photon. Energy of a photon is inversely proportional to its wavelength.

Complete answer:
Energy required to remove proton from its nucleus $ = 1MeV = {10^6}eV$
If $\lambda $ is the wavelength of the photon needed to remove the proton, then its energy (in Joules) is given by following relation,
$ \Rightarrow E = \dfrac{{hc}}{\lambda }$, where $h = 6.63 \times {10^{ - 34}}Js$ is Planck’s constant and $c = 3 \times {10^8}m/s$ is the speed of electromagnetic radiation in vacuum.
But, $1eV = 1.66 \times {10^{ - 19}}J$.
To remove the nucleus from its proton, the photon should carry energy equal to one required as mentioned before. Therefore, substituting the values of the constants, we get,
$
   \Rightarrow E = \dfrac{{hc}}{\lambda } = {10^6} \times (1.66 \times {10^{ - 19}}) \\
   \Rightarrow \lambda = \dfrac{{hc}}{{1.66 \times {{10}^{ - 13}}}} = \dfrac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.66 \times {{10}^{ - 13}}}} \\
   \Rightarrow \lambda = 1.2 \times {10^{ - 12}}m \\
$
Now, if we check the options then we find that the units are in $nm = {10^{ - 9}}m$.
So, $\lambda = 1.2 \times {10^{ - 9}} \times {10^{ - 3}}m = 1.2 \times {10^{ - 3}}nm$.
Comparing the options with the final result, we can conclude that option B is correct.

Note: Generally, the value of constants are provided in most of the examinations you will come across in future, but it is still advisable to remember some constants, like the ones used above along with the units. Be careful with powers of 10 while multiplication and division.