MHT CET · Physics · Dual Nature of Matter
A proton and alpha particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength of proton to that of alpha will be (mass of alpha particle is four times mass of proton.)
- A \(1: 2\)
- B \(2 \sqrt{2}: 1\)
- C \(1: 1\)
- D \(2: 1\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{2}: 1\)
Step-by-step Solution
Detailed explanation
De-Broglie wavelength is given by
\(
\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}
\)
If \(\lambda_1\) and \(\lambda_2\) are de-Broglie wavelengths of proton and alpha particle then
\(
\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{\mathrm{m}_2 \mathrm{q}_2}{\mathrm{~m}_1 \mathrm{q}_1}}=\sqrt{4 \times 2}=\sqrt{8}=2 \sqrt{2}
\)
\(
\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}
\)
If \(\lambda_1\) and \(\lambda_2\) are de-Broglie wavelengths of proton and alpha particle then
\(
\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{\mathrm{m}_2 \mathrm{q}_2}{\mathrm{~m}_1 \mathrm{q}_1}}=\sqrt{4 \times 2}=\sqrt{8}=2 \sqrt{2}
\)
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