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The velocity of the proton is one-fourth the velocity of the electron. What is the ratio of the de-Broglie wavelength of an electron to that of a proton?
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2765 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{4}$
The de-Broglie wavelength is given by
$$
\begin{gathered}
\lambda=\frac{h}{p}=\frac{h}{m v} \\
\Rightarrow \quad \frac{\lambda_{e}}{\lambda_{p}}=\frac{\frac{h}{m v_{e}}}{\frac{h}{m v_{p}}}=\frac{v_{p}}{v_{e}}=\frac{\frac{v_{e}}{4}}{v_{e}}=\frac{1}{4} \quad\left(\therefore v_{p}=\frac{1}{4} \times v_{e}\right)
\end{gathered}
$$
$$
\begin{gathered}
\lambda=\frac{h}{p}=\frac{h}{m v} \\
\Rightarrow \quad \frac{\lambda_{e}}{\lambda_{p}}=\frac{\frac{h}{m v_{e}}}{\frac{h}{m v_{p}}}=\frac{v_{p}}{v_{e}}=\frac{\frac{v_{e}}{4}}{v_{e}}=\frac{1}{4} \quad\left(\therefore v_{p}=\frac{1}{4} \times v_{e}\right)
\end{gathered}
$$
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