Discussion

The wavelength of 10 MeV proton is nearly equal to : (1F = 10\(^{-15}\) meter)

Answer: B

The wavelength is nearly 9 F.

A 10 MeV proton is non-relativistic (its rest energy is 938 MeV), so its momentum is \(p = \sqrt{2mK}\):

\[ pc = \sqrt{2 \times 938 \times 10} \approx 137\ \text{MeV} \]

\[ \begin{aligned} \lambda &= \frac{hc}{pc} = \frac{1240\ \text{MeV fm}}{137\ \text{MeV}} \\ &\approx 9.0\ \text{fm} \end{aligned} \]

Since \(\lambda \propto 1/\sqrt{K}\), the other values belong to other energies: 7 F to about 17 MeV, 11 F to about 7 MeV and 13 F to about 5 MeV.

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