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Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. What should be the kinetic energy of an electron for it to be able to probe a nucleon? Assume the diameter of a nucleon to be approximately $10^{-15} \mathrm{~m}$.
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To detect the properties of nucleons inside the nucleus the wavelength of particle which may detect nucleons or each particle (neutron and proton) present inside the nucleus is called nucleon.
Given, wavelength $\lambda=10^{-15} \mathrm{~m}$
To detect separate parts inside a nucleon the electron must have wavelength less than $10^{-15} \mathrm{~m}$.
We know that energy of free electron
$$
\mathrm{KE}=\mathrm{PE}
$$
Energy $\mathrm{E}=\mathrm{hv}=\frac{\mathrm{hc}}{\lambda}$
$$
=\frac{\mathrm{hc}}{\lambda}
$$
From Eq. (i) and Eq. (ii),
kinetic energy of electron
$$
\begin{aligned}
\mathrm{PE} &=\frac{\mathrm{hc}}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{10^{-15} \times 1.6 \times 10^{-19}} \mathrm{eV} \\
&=1.24 \times 10^9 \mathrm{eV} \\
\mathrm{KE} &=1.24 \times 10^9 \mathrm{eV}
\end{aligned}
$$
Given, wavelength $\lambda=10^{-15} \mathrm{~m}$
To detect separate parts inside a nucleon the electron must have wavelength less than $10^{-15} \mathrm{~m}$.
We know that energy of free electron
$$
\mathrm{KE}=\mathrm{PE}
$$
Energy $\mathrm{E}=\mathrm{hv}=\frac{\mathrm{hc}}{\lambda}$
$$
=\frac{\mathrm{hc}}{\lambda}
$$
From Eq. (i) and Eq. (ii),
kinetic energy of electron
$$
\begin{aligned}
\mathrm{PE} &=\frac{\mathrm{hc}}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{10^{-15} \times 1.6 \times 10^{-19}} \mathrm{eV} \\
&=1.24 \times 10^9 \mathrm{eV} \\
\mathrm{KE} &=1.24 \times 10^9 \mathrm{eV}
\end{aligned}
$$
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