Question 223: Dual Nature of Matter and Radiation

Question

A particle having electric charge $3 	imes 10^{-19} \mathrm{C}$ and mass $6 	imes 10^{-27} \mathrm{~kg}$ is accelerated by applying an electric potential of 1.21 V .
Wavelength of the matter wave associated with the particle is $\alpha 	imes 10^{-12} \mathrm{~m}$. The value of $\alpha$ is $\_\_\_\_$ .
(Take Planck's constant $=6.6 	imes 10^{-34} \mathrm{~J} . \mathrm{s}$ )

Accepted Answer

When a particle of charge $q$ is accelerated through a potential difference $V$, the work done on the particle is converted into its kinetic energy (K):
$$ \mathrm{K}=\mathrm{qV} $$
Given:
Charge $\mathrm{q}=3 	imes 10^{-19} \mathrm{C}$
Potential $\mathrm{V}=1.21 \mathrm{~V}$
$$ K=\left(3 	imes 10^{-19}ight) 	imes(1.21)=3.63 	imes 10^{-19} \mathrm{~J} $$
The momentum ( p ) of a particle of mass m is related to its kinetic energy ( K ) by:
$$ K=\frac{p^2}{2 m} \Rightarrow p=\sqrt{2 m K} $$
Substituting $\mathrm{K}=\mathrm{qV}$ :
$$ \mathrm{p}=\sqrt{2 \mathrm{mqV}} $$
The de-Broglie wavelength is given by:
$$ \lambda=\frac{\mathrm{h}}{\mathrm{p}} $$
Substituting the expression for momentum:
$$ \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}} $$
Substituting the given values:
$$ \mathrm{h}=6.6 	imes 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{~m}=6 	imes 10^{-27} \mathrm{~kg}, \mathrm{q}=3 	imes 10^{-19} \mathrm{C} 	ext { and } \mathrm{V}=1.21 \mathrm{~V} $$
So, the de-Broglie wavelength associated with the particle is,
$$ \lambda=\frac{6.6 	imes 10^{-34}}{\sqrt{2 	imes\left(6 	imes 10^{-27}ight) 	imes\left(3 	imes 10^{-19}ight) 	imes 1.21}} $$
$$\Rightarrow $$ $\lambda=\frac{6.6 	imes 10^{-34}}{\sqrt{36 	imes 10^{-46} 	imes 1.21}}$
$$\Rightarrow $$ $\lambda=\frac{6.6 	imes 10^{-34}}{6 	imes 10^{-23} 	imes 1.1}$
$$\Rightarrow $$ $\lambda=\frac{6.6 	imes 10^{-34}}{6.6 	imes 10^{-23}}$
$$\Rightarrow $$ $\lambda=1 	imes 10^{-11} \mathrm{~m}=\alpha 	imes 10^{-12} \mathrm{~m}$
$$\Rightarrow $$ $$ \lambda=10 	imes 10^{-12}=\alpha 	imes 10^{-12} \Rightarrow \alpha=10 $$
Therefore, the value of $\alpha$ is 10 .
Hence, the correct answer is $\mathbf{1 0}$.

Dual Nature of Matter and Radiation

Integer / Numeric Answer Type