Question

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. $1.2\,nm$
• B. $1.2 \times 10^{-3}\,nm$
• C. $1.2 \times 10^{-6}\,nm$
• D. $1.2 \times 10^1\,nm$

Answer 1

Answer: (B)

Here, $E = 1 \,MeV = 10^{6}eV$,
$ h = 6.63 \times10^{-34} \,J\, s$,
$ c= 3\times10^{8} \,m \,s^{-1} $
$ \therefore hc = \frac{6.63 \times 10^{-34} \times 3\times 10^{8}}{1.6 \times 10^{-9}} \approx 1240 \,eV \,nm$
As $E =\frac{hc}{\lambda}$
or $\lambda = \frac{hc}{E} =\frac{ 1240\, eV \,nm }{10^{6} eV}$
$ = 1.24 \times 10^{-3}\, nm$