Question

An $\alpha$-particle is accelerated through a potential difference of $V$ volts from rest. The de-Broglie wavelength associated with it is (in Angstrom)

Answer 1

Step 1: Given data

$$\begin{gathered} m_{\alpha }=4m_p \\ {m}_p=1.67\times {10}^{-27}kg \\ {q}_{\alpha }=2e=2\times 1.6\times {10}^{-19}C \\ h=6.626\times {10}^{-34} \end{gathered}$$

$pismomentum$

where $m_{\alpha }$ is the mass of $\alpha$-particle, $m_p$ is the mass of proton, $q_{\alpha }$ is the charge on alpha particle, $e$ is the electronic charge and $h$ is the planks constant.

Step 2: To find

The de-Broglie wavelength associated with the $\alpha$-particle

Step 3: Calculation

The kinetic energy gained by the $\alpha$-particle by potential difference $V$ is given by,

$$\begin{gathered} K_{\alpha }=\frac{p^2}{2m_{\alpha }} \\ {K}_{\alpha }=qV \\ \Rightarrow \frac{p^2}{2m_{\alpha }}=q_{\alpha }V \end{gathered}$$

Substituting the given values,

$$\begin{gathered} \Rightarrow \frac{p^2}{2m_{\alpha }}=q_{\alpha }V \\ \Rightarrow p^2=q_{\alpha }V\times 2m_{\alpha } \\ \Rightarrow p^2=2e\times V\times 2\times 4m_p \\ \Rightarrow p^2=16\times eV{m}_p \\ \Rightarrow p=4\times \sqrt{1.6\times {10}^{-19}\times V\times 1.67\times {10}^{-27}} \\ \Rightarrow p=4\times 1.634\times {10}^{-23}\times \sqrt{V} \end{gathered}$$

de-Broglie wavelength,

$$\begin{gathered} \lambda =\frac{h}{p} \\ \Rightarrow \lambda =\frac{6.626\times {10}^{-34}}{4\times 1.634\times {10}^{-23}\times \sqrt{V}} \\ \Rightarrow \lambda =\frac{0.101}{\sqrt{V}}A° \end{gathered}$$

The de-Broglie wavelength associated with the $\alpha$-particle is $\frac{0.101}{\sqrt{V}}A°$.