Solve this following
Question:

An electron, a doubly ionized helium ion $\left(\mathrm{He}^{++}\right)$and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths $\lambda_{e}, \lambda_{\mathrm{He}^{+}}$and $\lambda_{\mathrm{P}}$ is:

  1. $\lambda_{e}<\lambda_{P}<\lambda_{H e^{++}}$

  2. $\lambda_{\mathrm{e}}<\lambda_{\mathrm{He}^{+}}=\lambda_{\mathrm{P}}$

  3. $\lambda_{\mathrm{e}}>\lambda_{\mathrm{He}^{++}}>\lambda_{\mathrm{P}}$

  4. $\lambda_{\mathrm{e}}>\lambda_{\mathrm{P}}>\lambda_{\mathrm{He}^{++}}$


Correct Option: , 4

Solution:

$\lambda=\frac{\mathrm{h}}{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}$

$\lambda \propto \frac{1}{\sqrt{\mathrm{m}}} \Rightarrow \lambda=\frac{\mathrm{C}}{\sqrt{\mathrm{m}}}$

$\mathrm{m}_{\mathrm{He}^{++}}>\mathrm{m}_{\mathrm{P}}>\mathrm{m}_{\mathrm{e}}$

$\therefore \lambda_{\mathrm{He}^{++}}<\lambda_{\mathrm{P}}<\lambda_{\mathrm{e}}$

$\therefore$ correct option is (4)

 

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