MHT CET · Physics · Dual Nature of Matter
An electron accelerated through a potential difference ' \(V_1\) ' has a de-Broglie wavelength ' \(\lambda\) '. When the potential is changed to ' \(\mathrm{V}_2\) ' its de-Broglie wavelength increases by \(50 \%\). The value of \(\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)\) is
- A 3:1
- B 9:4
- C 3:2
- D 4:1
Answer & Solution
Correct Answer
(B) 9:4
Step-by-step Solution
Detailed explanation
For electron, de Broglie wavelength, \(\lambda=\frac{1.228}{\sqrt{\mathrm{V}}}\) Given: \(\lambda_2=\lambda_1+0.5 \lambda_1=1.5 \lambda_1\)
\(\begin{aligned}
& \therefore \quad \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{V_1}{V_2}} \\
& \quad \Rightarrow \frac{V_1}{V_2}=\left(\frac{\lambda_2}{\lambda_1}\right)^2=\left(\frac{1.5 \lambda_1}{\lambda_1}\right)^2=\left(\frac{3}{2}\right)^2
\end{aligned}\)
\(=\frac{9}{4}\)
\(\begin{aligned}
& \therefore \quad \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{V_1}{V_2}} \\
& \quad \Rightarrow \frac{V_1}{V_2}=\left(\frac{\lambda_2}{\lambda_1}\right)^2=\left(\frac{1.5 \lambda_1}{\lambda_1}\right)^2=\left(\frac{3}{2}\right)^2
\end{aligned}\)
\(=\frac{9}{4}\)
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