If a proton and $\alpha$ -particle are accelerated through the same potential difference, the ratio of…
If a proton and $\alpha$ -particle are accelerated through the same potential difference, the ratio of deBroglie wavelengths $\lambda_{\mathrm{p}}$ and $\lambda_{\alpha}$ is
3
$2 \sqrt{2}$
1
2
Solution
Key Point : The de-Broglie wavelength of a particle of mass m and moving with velocity \(v\) is given by
\(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}(\because \mathrm{p}=\mathrm{mv})\)
de-Broglie wavelength of a proton of mass \(m_1\) and kinetic energy k is given by
\(\begin{aligned}
& \lambda_1=\frac{h}{\sqrt{2 m_1 k}}(\because p=\sqrt{2 m k}) \\
& \lambda_1=\frac{h}{\sqrt{2 m_1 q V}} \cdots \text { (i) }[\because k=q V]
\end{aligned}\)
For an alpha particle mass \(\mathrm{m}_2\) carrying charge \(\mathrm{q}_0\) is accelerated through potential V , then
\(\lambda_2=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_2 \mathrm{q}_0 \mathrm{~V}}}\)
\(\because\) For \(\alpha\) - particle \(\left({ }_2^4 \mathrm{He}ight): q_0=2 \mathrm{q}\) and \(\mathrm{m}_2=4 \mathrm{~m}_1\)
\(\therefore \lambda_2=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_1 \times 2 \mathrm{q} \times \mathrm{V}}} \ldots (ii)\)
The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get \(\frac{\lambda_1}{\lambda_2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_1 \mathrm{qV}}} \times \frac{\sqrt{2 \times \mathrm{m}_1 \times 4 \times 2 \mathrm{qV}}}{\mathrm{h}}=\frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\)
We get \(\frac{\lambda_1}{\lambda_2}=2 \sqrt{2}\)