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
- A \(\frac{\lambda}{3} \sqrt{\frac{M}{m}}\)
- B \(\frac{\lambda}{3} \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}\)
- C \(\frac{\lambda}{6} \sqrt{\frac{m}{M}}\)
- D \(\frac{\lambda}{6} \sqrt{\frac{M}{m}}\)
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,
\(\begin{aligned}
& \lambda_p=\frac{h}{\sqrt{2 q(9 V) M}} \\
& \therefore \frac{\lambda}{\lambda_p}=3 \sqrt{\frac{M}{m}} \\
& \Rightarrow \lambda_p=\frac{\lambda}{3} \sqrt{\frac{m}{M}}
\end{aligned}\)

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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