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A charged particle is accelerated from rest through a certain potential difference. The de-Broglie wavelength is $\lambda_1$ when it is accelerated through $V_1$ and is $\lambda_2$ when accelerated through $V_2$. The ratio $\lambda_1 / \lambda_2$ is
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Verified Answer
The correct answer is:
$V_2^{1 / 2}: V_1^{1 / 2}$
The de-Broglie wavelength associated with a charged particle is given by
$$
\lambda=\frac{h}{\sqrt{m q V}}
$$
where, $h=$ planck's constant
$m=$ mass of charged particle
$q=$ charge on the particle
$V=$ potential at which the charged particle is accelerated.
$\therefore$
$$
\lambda_1=\frac{h}{\sqrt{m_1 q V_1}}, \lambda_2=\frac{h}{\sqrt{m_2 q V_2}}
$$
Here,
$$
\begin{aligned}
m_1 & =m_2 \\
\frac{\lambda_1}{\lambda_2} & =\sqrt{\frac{m_2 q V_2}{m_1 q V_1}}=\sqrt{\frac{V_2}{V_1}} \\
& =V_2^{\frac{1}{2}} \cdot V_1^{\frac{1}{2}}
\end{aligned}
$$
$$
\lambda=\frac{h}{\sqrt{m q V}}
$$
where, $h=$ planck's constant
$m=$ mass of charged particle
$q=$ charge on the particle
$V=$ potential at which the charged particle is accelerated.
$\therefore$
$$
\lambda_1=\frac{h}{\sqrt{m_1 q V_1}}, \lambda_2=\frac{h}{\sqrt{m_2 q V_2}}
$$
Here,
$$
\begin{aligned}
m_1 & =m_2 \\
\frac{\lambda_1}{\lambda_2} & =\sqrt{\frac{m_2 q V_2}{m_1 q V_1}}=\sqrt{\frac{V_2}{V_1}} \\
& =V_2^{\frac{1}{2}} \cdot V_1^{\frac{1}{2}}
\end{aligned}
$$
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