JEE Main 2023 — Structure of Atom Question with Solution

Given:
Kinetic energy(K.E.)=$4.55\times {10}^{-29}\mathrm{J}$
Mass of electron$$\begin{gathered} \left({\mathrm{m}}_{\mathrm{e}}\right)=9.1\times {10}^{-31}\mathrm{kg} \\ \mathrm{h}=6.6\times {10}^{-34}\mathrm{J}-\sec \end{gathered}$$

debroglie wavelength ${\mathrm{\lambda }}_{\mathrm{d}}=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2\mathrm{mKE}}}$

Here $\mathrm{\lambda }$ is the de Broglie wavelength, $\mathrm{h}$ is Planck's constant, $\mathrm{m}$is the mass of the particle, $\mathrm{v}$ is its velocity, and $\mathrm{p}$ is the momentum of the particle, which is equal to $\mathrm{mv}$.

$\lambda =\frac{\mathrm{h}}{\sqrt{2\mathrm{mKE}}}$

$=\frac{6.6\times {10}^{-34}}{\sqrt{2\times 9\times {10}^{-31}\times 4.5\times {10}^{-29}}}$$=\frac{6.6\times {10}^{-34}}{9\times {10}^{-30}}=\frac{6.6}{9}\times {10}^{-4}$

$=\frac{66}{9}\times {10}^{-5}$

$=7.33\times {10}^{-5}$