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MHT CET · Physics · Dual Nature of Matter

According to de-Broglie hypothesis if an electron of mass ' \(\mathrm{m}\) ' is accelerated by potential difference ' \(\mathrm{V}\) ' then associated wavelength is ' \(\lambda\) '. When a proton of mass ' \(M\) ' is accelerated through potential difference ' \(9 \mathrm{~V}\) ' then the wavelength associated with it is

  1. A \(\frac{\lambda}{3} \sqrt{\frac{M}{m}}\)
  2. B \(\frac{\lambda}{3} \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}\)
  3. C \(\frac{\lambda}{6} \sqrt{\frac{m}{M}}\)
  4. D \(\frac{\lambda}{6} \sqrt{\frac{M}{m}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{\lambda}{3} \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}\)

Step-by-step Solution

Detailed explanation

Do Broglie relation

Energy conservation,

Energy vs momentum relation

Using equation (1), (2) and (3)
\(\begin{aligned}
& \lambda=\frac{h}{p}=\frac{h}{\sqrt{2 K m}} \\
& \lambda=\frac{h}{\sqrt{2 q V m}}
\end{aligned}\)
Now, for proton of mass \(M\) that is accelerated through \(9 \mathrm{~V}\), the de-broglie wavelength can be written as,
\(\begin{aligned}
& \lambda_p=\frac{h}{\sqrt{2 q(9 V) M}} \\
& \therefore \frac{\lambda}{\lambda_p}=3 \sqrt{\frac{M}{m}} \\
& \Rightarrow \lambda_p=\frac{\lambda}{3} \sqrt{\frac{m}{M}}
\end{aligned}\)
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