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The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nuclear with $1 \mathrm{MeV}$ energy is nearly
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The correct answer is:
$1.2 \times 10^{-3} \mathrm{~nm}$
$1.2 \times 10^{-3} \mathrm{~nm}$
Energy of a photon is $E=\frac{h c}{\lambda}$
Where $\lambda$ is the minimum wavelength of the photon required to eject the proton from nucleus.
Energy of photon must be equal to the binding energy of proton. So, energy of a photon, $\mathrm{E}=1 \mathrm{MeV} \Rightarrow 10^6 \mathrm{eV} \quad$ (given)
Now, $\left(E=\frac{h c}{\lambda}\right)$
So, $\lambda=\frac{\mathrm{hc}}{\mathrm{E}}=\left(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{10^6 \mathrm{eV}}\right)$
$\Rightarrow \quad$ So, $\lambda=\frac{\mathrm{hc}}{\mathrm{E}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{10^6 \times 1.6 \times 10^{-19} \mathrm{~J}}$
$=1.24 \times 10^{-9} \times 10^{-3}$
$=1.24 \times 10^{-3} \mathrm{~nm}$
Where $\lambda$ is the minimum wavelength of the photon required to eject the proton from nucleus.
Energy of photon must be equal to the binding energy of proton. So, energy of a photon, $\mathrm{E}=1 \mathrm{MeV} \Rightarrow 10^6 \mathrm{eV} \quad$ (given)
Now, $\left(E=\frac{h c}{\lambda}\right)$
So, $\lambda=\frac{\mathrm{hc}}{\mathrm{E}}=\left(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{10^6 \mathrm{eV}}\right)$
$\Rightarrow \quad$ So, $\lambda=\frac{\mathrm{hc}}{\mathrm{E}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{10^6 \times 1.6 \times 10^{-19} \mathrm{~J}}$
$=1.24 \times 10^{-9} \times 10^{-3}$
$=1.24 \times 10^{-3} \mathrm{~nm}$
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