Question
An electron and photon have same energy 100eV. Which has greater associated wavelength?
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Answer
De-Broglie wavelength associated with electron, $\lambda_\text{e}=\frac{\text{h}}{\sqrt{2\text{mE}_\text{e}}}\Rightarrow\ \text{E}_\text{}e=\frac{\text{h}^2}{2\text{m}\lambda^2_\text{e}}\dots(\text{i})$ Also wavelength of photon of energy $E_{ph}$ is,$\text{E}_\text{ph}=\frac{\text{hc}}{\lambda_{\text{ph}}}\Rightarrow\ \text{E}^2_{\text{ph}}=\frac{\text{h}^2\text{c}^2}{\lambda^2_\text{ph}}\dots(\text{ii})$
Given $E_e = E_{ph} = E (say) = 100eV$ Dividing (ii) by (i) and using (iii), we get $\text{E}=\frac{\frac{\text{h}^2\text{c}^2}{\lambda^2_\text{ph}}}{\frac{\text{h}^2}{2\text{m}\lambda^2_\text{e}}}$ or $\text{E}=\frac{2\text{mc}^2\lambda^2_\text{e}}{\lambda^2_{\text{ph}}}$ $\therefore\ \frac{\lambda_\text{e}}{\lambda_{\text{ph}}}=\sqrt{\frac{\text{E}}{2\text{mc}^2}}$ As $\text{E}=100\text{eV } 2\text{mc}^2\cong1\text{MeV}$ $\therefore\ \text{E}<<2\text{mc}^2\Rightarrow\ \lambda_\text{e}<\lambda_\text{ph}$ That is, wavelength associated with photon is greater as compared to electron of same energy.
Given $E_e = E_{ph} = E (say) = 100eV$ Dividing (ii) by (i) and using (iii), we get $\text{E}=\frac{\frac{\text{h}^2\text{c}^2}{\lambda^2_\text{ph}}}{\frac{\text{h}^2}{2\text{m}\lambda^2_\text{e}}}$ or $\text{E}=\frac{2\text{mc}^2\lambda^2_\text{e}}{\lambda^2_{\text{ph}}}$ $\therefore\ \frac{\lambda_\text{e}}{\lambda_{\text{ph}}}=\sqrt{\frac{\text{E}}{2\text{mc}^2}}$ As $\text{E}=100\text{eV } 2\text{mc}^2\cong1\text{MeV}$ $\therefore\ \text{E}<<2\text{mc}^2\Rightarrow\ \lambda_\text{e}<\lambda_\text{ph}$ That is, wavelength associated with photon is greater as compared to electron of same energy.
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