The velocity of particle $\mathrm{A}$ is $0.1 \mathrm{~ms}^{-1}$ and that of particle $\mathrm{B}$ is $0.05…
- $2: 5$
- $3: 4$
- $6: 4$
- $5: 2$
Solution
$\mathrm{m}_{\mathrm{B}}=5 \mathrm{~m}_{\mathrm{A}}$
de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$
$\therefore \frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{h} / \mathrm{m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}}{\mathrm{h} / \mathrm{m}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}}=\frac{\mathrm{m}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}}{\mathrm{m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}}$
$=\frac{5 \mathrm{~m}_{\mathrm{A}} \times 0.05}{\mathrm{~m}_{\mathrm{A}} \times 0.1}=5 \times 0.5=2.5=5 / 2$
$\therefore \lambda_{\mathrm{A}}: \lambda_{\mathrm{B}}=5: 2$ .
Asked in: JEE-TOPICTESTS-CHEMISTRY