Question 318: Dual Nature of Matter and Radiation

Question

An electron accelerated through a potential difference '$V_1$' has a de-Broglie wavelength '$\lambda$'. When the potential is changed to '$V_2$' its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)$ is

Accepted Answer

For electron, de Broglie wavelength, $\lambda=\frac{1.228}{\sqrt{V}}$
Given: $\lambda_2=\lambda_1+0.5 \lambda_1=1.5 \lambda_1$
$\begin{array}{ll}
	herefore \quad & \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{V_1}{V_2}} \
& \Rightarrow \frac{V_1}{V_2}=\left(\frac{\lambda_2}{\lambda_1}ight)^2=\left(\frac{1.5 \lambda_1}{\lambda_1}ight)^2=\left(\frac{3}{2}ight)^2=\frac{9}{4}
\end{array}$

Answer

$3: 1$

Answer

$9: 4$

Answer

$3: 2$

Answer

$4: 1$

Dual Nature of Matter and Radiation

An electron accelerated through a potential difference 'V_1' has a de-Broglie wavelength '\lambda'. When the potential is changed to 'V_2' its de-Broglie wavelength increases by 50 \%. The value of \left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right) is

Choose the correct answer: