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Two particles of masses $m$ and $2 m$ have equal kinetic energies. The de-Broglie wavelengths are in the ratio of
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Verified Answer
The correct answer is:
$\sqrt{2}: 1$
As we know
$$
\begin{aligned}
& \lambda=\frac{h}{m v} \\
& \lambda=\frac{h}{\sqrt{2 m E}}
\end{aligned}
$$
$$
\left[\because E=\frac{1}{2} m v^2\right]
$$
Since, $h$ is a constant and kinetic energy is same for both particles, so,
$$
\begin{aligned}
\lambda & =\frac{1}{\sqrt{m}} \Rightarrow \lambda_1=\frac{1}{\sqrt{m}} \\
\text { and } \quad \lambda_2 & =\frac{1}{\sqrt{2 m}}
\end{aligned}
$$
The ratio will be
$$
\frac{\lambda_1}{\lambda_2}=\frac{\sqrt{2 m}}{\sqrt{m}}=\frac{\sqrt{2}}{1}
$$
The ratio of de-Broglie wavelength is $\sqrt{2}: 1$.
$$
\begin{aligned}
& \lambda=\frac{h}{m v} \\
& \lambda=\frac{h}{\sqrt{2 m E}}
\end{aligned}
$$
$$
\left[\because E=\frac{1}{2} m v^2\right]
$$
Since, $h$ is a constant and kinetic energy is same for both particles, so,
$$
\begin{aligned}
\lambda & =\frac{1}{\sqrt{m}} \Rightarrow \lambda_1=\frac{1}{\sqrt{m}} \\
\text { and } \quad \lambda_2 & =\frac{1}{\sqrt{2 m}}
\end{aligned}
$$
The ratio will be
$$
\frac{\lambda_1}{\lambda_2}=\frac{\sqrt{2 m}}{\sqrt{m}}=\frac{\sqrt{2}}{1}
$$
The ratio of de-Broglie wavelength is $\sqrt{2}: 1$.
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