Question 221: Dual Nature of Matter and Radiation

Question

The de Broglie wavelength for an electron accelerated through the potential difference of $V_1$ volt is $\lambda_1$. When the potential difference is changed to $V_2$ volt, the associated de Broglie wavelength is increased by $50 \%$. If $\left(V_1 / V_2\right)=(9 / \alpha)$, then the value of $\alpha$ is $\_\_\_\_$。

Accepted Answer

The de Broglie wavelength $\lambda$ is given by:
$$ \lambda=\frac{\mathrm{h}}{\mathrm{p}} $$
Where h is Planck's constant and p is the momentum.
The kinetic energy ( $K$ ) of an electron accelerated through a potential $V$ is $K=e V=\frac{p^2}{2 m}$
$$ \mathrm{p}=\sqrt{2 \mathrm{meV}} $$
Substituting p into the wavelength formula:
$$ \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}} $$
For a given particle (electron), the wavelength is inversely proportional to the square root of the potential,
$$ \lambda \propto \frac{1}{\sqrt{V}} $$
So, the ratio of wavelengths in two cases is,
$$ \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{V_1}{V_2}} $$
The initial wavelength is $\lambda_1$.
The final wavelength is $\lambda_2=\lambda_1+50 \%$ of $\lambda_1=1.5 \lambda_1=\frac{3}{2} \lambda_1$.
Substituting these into the ratio :
$$ \frac{\frac{3}{2} \lambda_1}{\lambda_1}=\sqrt{\frac{V_1}{V_2}} $$
$$\Rightarrow $$ $\frac{3}{2}=\sqrt{\frac{V_1}{V_2}}$
$$\Rightarrow $$ $\left(\frac{3}{2}\right)^2=\frac{V_1}{V_2}$
$$\Rightarrow $$ $$ \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{9}{4}=\frac{9}{\alpha} \Rightarrow \alpha=4 $$
Therefore, the value of $\alpha$ is 4 .

Dual Nature of Matter and Radiation

Integer / Numeric Answer Type