Question 32: Dual Nature of Matter and Radiation

Question

An electron is travelling with a velocity $v$ in free space and when it enters a medium, its velocity is reduced by $20 \%$. The de Broglie wavelength of electron in the medium is $\alpha \lambda_0$, where $\lambda_0$ is its de Broglie wavelength in free space. The value of $\alpha$ is $\_\_\_\_$ .

Accepted Answer

The de Broglie wavelength ( $\lambda$ ) of a particle with mass $m$ and velocity $v$ is given by:
$$ \lambda=\frac{\mathrm{h}}{\mathrm{mv}} $$
Where h is Planck's constant.
For a particle of constant mass (like an electron), the wavelength is inversely proportional to its velocity:
$$ \lambda \propto \frac{1}{v} $$
Let the initial velocity in free space be $\mathrm{v}_0$.
The velocity in the medium ( $\mathrm{v}_{\mathrm{m}}$ ) is reduced by $20 \%$ :
$$ \mathrm{v}_{\mathrm{m}}=\mathrm{v}_0-20 \% 	ext { of } \mathrm{v}_0=\mathrm{v}_0-0.20 \mathrm{v}_0=0.80 \mathrm{v}_0 $$
Let $\lambda_0$ be the wavelength in free space and $\lambda_{\mathrm{m}}$ be the wavelength in the medium.
$$ \frac{\lambda_{\mathrm{m}}}{\lambda_0}=\frac{\mathrm{v}_0}{\mathrm{v}_{\mathrm{m}}} $$
Substituting the value of $\mathrm{v}_{\mathrm{m}}$ :
$\frac{\lambda_{\mathrm{m}}}{\lambda_0}=\frac{\mathrm{v}_0}{0.80 \mathrm{v}_0}=\frac{1}{0.80}$
$$\Rightarrow $$ $\frac{\lambda_{\mathrm{m}}}{\lambda_0}=\frac{100}{80}=1.25$
$$\Rightarrow $$ $\lambda=1.25 \lambda_0=\alpha \lambda_0$
$$\Rightarrow $$$$ \alpha=1.25 $$
Therefore, the correct option is (C).

Answer

1.20

Answer

1.0

Answer

1.25

Answer

0.75

Dual Nature of Matter and Radiation

An electron is travelling with a velocity v in free space and when it enters a medium, its velocity is reduced by 20 \%. The de Broglie wavelength of electron in the medium is \alpha \lambda_0, where \lambda_0 is its de Broglie wavelength in free space. The value of \alpha is \_\_\_\_ .

Choose the correct answer: