Question

The wavelength of a microscopic particle of mass $9.1 \times 10^{-31} kg$ is $182\, nm$, its kinetic energy in $J$ is $\left(h=6.625 \times 10^{-34}\, J s \right)$

• A. $7.28 \times 10^{-23}$
• B. $7.28 \times 10^{-24}$
• C. $3.64 \times 10^{-23}$
• D. $3.64 \times 10^{-24}$

Answer 1

Answer: (B)

Given, mass of particles $=9.1 \times 10^{-31} \,kg$

Wavelength $(\lambda)=182\, nm =182 \times 10^{-9} \,m$

According to de-Broglie equation,

$\lambda =\frac{h}{m v}$

$ \Rightarrow v=\frac{h}{m \times \lambda}$

$v =\frac{6.625 \times 10^{-34} Js }{9.1 \times 10^{-31} kg \times 182 \times 10^{-9}\, m } $

$=0.004 \times 10^{6} \,m / s =4 \times 10^{3} \,m / s$

Therefore, kinetic energy,

$KE =\frac{1}{2} m v^{2}= \frac{1}{2} \times 9.1 \times 10^{-31} \times\left(4 \times 10^{3}\right)^{2} $

$=7.28 \times 10^{-24} \,J$