Question
A proton and an $\alpha -$particle are accelerated, using the same potential difference. How are the de$-$Broglie wavelengths $\lambda _p$ and $\lambda _a$ related to each other?
✓
Answer
Key concept:
Hence de$-$Broglie wavelength : $\lambda=\frac{\text{h}}{\text{p}}=\frac{\text{h}}{\sqrt{2\text{mE}}}=\frac{\text{h}}{\sqrt{2\text{mqV}}}$
In this problrm since both proton and $\alpha-$ particle are accelerated through same potential difference,
We know that, $\lambda=\frac{\text{h}}{\sqrt{2\text{mqv}}}$
$\therefore\ \lambda\propto\frac{1}{\sqrt{\text{mq}}}$
$\frac{\lambda_\text{p}}{\lambda_\alpha}=\frac{\sqrt{\text{m}_\alpha\text{q}_\alpha}}{\text{m}_\text{p}\text{q}_\text{p}}$
$=\frac{\sqrt{4\text{m}_\text{p}\times2\text{e}}}{\sqrt{\text{m}_\text{p}\times\text{e}}}$
$=\sqrt{8}$
$\therefore\ \lambda_\text{p}=\sqrt{8}\lambda_\alpha$
i.e., wavelength of proton is times wavelength of $\alpha-$particle.
Important point:
De$-$Broglie wavelength associated with the charged particles:
The energy of a charged particle acceletated through potential difference $V$ is $\text{E}=\frac{1}{2}\text{mv}^2=\text{qV}$
Hence de$-$Broglie wavelength $\lambda=\frac{\text{h}}{\text{p}}=\frac{\text{h}}{\sqrt{2\text{mE}}}=\frac{\text{h}}{\sqrt{2\text{mqV}}}$
$\lambda_\text{Electron}=\frac{12.27}{\sqrt{\text{V}}}\mathring{\text{A}},\lambda_\text{Proton}=\frac{0.286}{\sqrt{\text{V}}}\mathring{\text{A}}$
$\lambda_\text{Deutron}=\frac{0.200}{\sqrt{\text{V}}}\mathring{\text{A}},\lambda_{\alpha-\text{particle}}=\frac{0.101}{\sqrt{\text{V}}}\mathring{\text{A}}$
Hence de$-$Broglie wavelength : $\lambda=\frac{\text{h}}{\text{p}}=\frac{\text{h}}{\sqrt{2\text{mE}}}=\frac{\text{h}}{\sqrt{2\text{mqV}}}$
In this problrm since both proton and $\alpha-$ particle are accelerated through same potential difference,
We know that, $\lambda=\frac{\text{h}}{\sqrt{2\text{mqv}}}$
$\therefore\ \lambda\propto\frac{1}{\sqrt{\text{mq}}}$
$\frac{\lambda_\text{p}}{\lambda_\alpha}=\frac{\sqrt{\text{m}_\alpha\text{q}_\alpha}}{\text{m}_\text{p}\text{q}_\text{p}}$
$=\frac{\sqrt{4\text{m}_\text{p}\times2\text{e}}}{\sqrt{\text{m}_\text{p}\times\text{e}}}$
$=\sqrt{8}$
$\therefore\ \lambda_\text{p}=\sqrt{8}\lambda_\alpha$
i.e., wavelength of proton is times wavelength of $\alpha-$particle.
Important point:
De$-$Broglie wavelength associated with the charged particles:
The energy of a charged particle acceletated through potential difference $V$ is $\text{E}=\frac{1}{2}\text{mv}^2=\text{qV}$
Hence de$-$Broglie wavelength $\lambda=\frac{\text{h}}{\text{p}}=\frac{\text{h}}{\sqrt{2\text{mE}}}=\frac{\text{h}}{\sqrt{2\text{mqV}}}$
$\lambda_\text{Electron}=\frac{12.27}{\sqrt{\text{V}}}\mathring{\text{A}},\lambda_\text{Proton}=\frac{0.286}{\sqrt{\text{V}}}\mathring{\text{A}}$
$\lambda_\text{Deutron}=\frac{0.200}{\sqrt{\text{V}}}\mathring{\text{A}},\lambda_{\alpha-\text{particle}}=\frac{0.101}{\sqrt{\text{V}}}\mathring{\text{A}}$
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