Question 339: 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

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