KCET · Physics · Dual Nature of Matter
If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is
- A zero
- B less than that of a proton
- C more than that of a proton
- D equal to that of a proton
Answer & Solution
Correct Answer
(C) more than that of a proton
Step-by-step Solution
Detailed explanation
Given,
\[
\begin{aligned}
\lambda &=\frac{h}{m v}=\frac{h}{\sqrt{2 m K E}} \\
K E &=\frac{h^{2}}{2 m \lambda^{2}}
\end{aligned}
\]
\[
\Rightarrow \quad \mathrm{KE}=\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}
\]
As \(\lambda\) is same for both electron and proton
So, \(\quad \mathrm{KE} \propto \frac{1}{\mathrm{~m}}\)
Hence, kinetic energy will be maximum for particle with lesser mass, electron.
\[
\begin{aligned}
\lambda &=\frac{h}{m v}=\frac{h}{\sqrt{2 m K E}} \\
K E &=\frac{h^{2}}{2 m \lambda^{2}}
\end{aligned}
\]
\[
\Rightarrow \quad \mathrm{KE}=\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}
\]
As \(\lambda\) is same for both electron and proton
So, \(\quad \mathrm{KE} \propto \frac{1}{\mathrm{~m}}\)
Hence, kinetic energy will be maximum for particle with lesser mass, electron.
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