KCET · Physics · Atomic Physics
A particle is dropped from a height ' \( H \) '. The de Broglie wavelength of the particle depends on
height as
- A \( \mathrm{H} \)
- B \( \mathrm{H}^{0} \)
- C \( \mathrm{H}^{\frac{1}{2}} \)
- D \( \mathrm{H}^{-\frac{1}{2}} \)
Answer & Solution
Correct Answer
(D) \( \mathrm{H}^{-\frac{1}{2}} \)
Step-by-step Solution
Detailed explanation
According to the de Broglie equation the wavelength of the particle is given as
\[
\lambda=\frac{h}{m v}
\]
For a particle dropped from height \( \mathrm{H} \) is given as
\[
\begin{array}{l}
v=\sqrt{2 g H} \\
\lambda=\frac{h}{m \sqrt{g H} \times 2} \\
\Rightarrow \lambda \propto \frac{1}{\sqrt{H}}
\end{array}
\]
Therefore, the de Broglie wavelength of particle depends on height as \( H^{-\frac{1}{2}} \)
\[
\lambda=\frac{h}{m v}
\]
For a particle dropped from height \( \mathrm{H} \) is given as
\[
\begin{array}{l}
v=\sqrt{2 g H} \\
\lambda=\frac{h}{m \sqrt{g H} \times 2} \\
\Rightarrow \lambda \propto \frac{1}{\sqrt{H}}
\end{array}
\]
Therefore, the de Broglie wavelength of particle depends on height as \( H^{-\frac{1}{2}} \)
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