Question

What will be the kinetic energy of an electron having de-Broglie wavelength $2{A^ \circ }$?
A) $37.5eV$
B) $75eV$
C) $150eV$
D) $300eV$

Answer 1

In this question, we can calculate the kinetic energy by using the relation between the momentum and kinetic energy. We can change this relation into the relation between kinetic energy and de-Broglie wavelength as we have given the de-Broglie wavelength.

Complete step by step solution: -
We know that if the $p$ is the momentum of the electron and $\lambda$ is the de-Broglie wavelength then the relation between the momentum and the de-Broglie wavelength is given as-
$\lambda = \dfrac{h}{p}$
Where $h$ is the Planck’s constant.
Now, if $m$ is the mass of the electron then the kinetic energy K.E. is given as-
$K.E. = \dfrac{{{p^2}}}{{2m}}$
Or we can write this formula in terms of de-Broglie wavelength$\lambda$ as-
$K.E. = \dfrac{1}{{2m}}{\left( {\dfrac{h}{\lambda }} \right)^2} \\\ \Rightarrow K.E.Y. = \dfrac{{{h^2}}}{{2m{\lambda ^2}}} \\\$
Now, we know that Planck’s constant $h = 6.626 \times {10^{ - 34}}$ , mass of electron $m = 9.11 \times {10^{ - 31}}$ and the de-Broglie wavelength $\lambda = 2 \times {10^{ - 10}}m$ (as $1{A^ \circ } = {10^{ - 10}}m$). So, substituting these values in the above equation, we get-
$K.E. = \dfrac{{{{\left( {6.626 \times {{10}^{ - 34}}} \right)}^2}}}{{2 \times 9.11 \times {{10}^{ - 31}} \times {{\left( {2 \times {{10}^{ - 10}}} \right)}^2}}} \\\ \Rightarrow K.E. = \dfrac{{43.9039 \times {{10}^{ - 68}}}}{{18.22 \times {{10}^{ - 31}} \times 4 \times {{10}^{ - 20}}}} \\\ \Rightarrow K.E. = \dfrac{{43.9039 \times {{10}^{ - 68}}}}{{72.88 \times {{10}^{ - 31 - 20}}}} \\\ \Rightarrow K.E. = 0.6024 \times {10^{ - 68 + 31 + 20}} \\\ \Rightarrow K.E. = 0.6024 \times {10^{ - 17}}J \\\$
Now, we know that
If $1J = 6.2415 \times {10^{18}}eV$
Then

$$0.6024 \times {10^{ - 17}}J = 0.6024 \times {10^{ - 17}} \times 6.24150.6024 \times {10^{18}} \\\ \Rightarrow 0.6024 \times {10^{ - 17}}J = 3.7601 \times 10 \\\ \Rightarrow 0.6024 \times {10^{ - 17}}J = 37.601eV \\\$$

Hence, $37.5eV$($\sim 37.601eV$) will be the kinetic energy of an electron having de-Broglie wavelength $2{A^ \circ }$.

Therefore, option A is correct.

Additional Information: - According to the dual nature of particles, the de-Broglie wavelength is a wavelength which is attached in all the particles in a way which determines the probability density of finding the particle at a given point of the space.

Note: - In this question, we have to remember the relation between kinetic energy and momentum of the electron. According to the question, we have the de-Broglie wavelength. So, we have to change the relation between kinetic energy and the momentum into the kinetic energy and de-Broglie wavelength by substituting $p = \dfrac{h}{\lambda }$ in $K.E. = \dfrac{{{p^2}}}{{2m}}$.