If the kinetic energy of a particle is increased to 16 times its previous value, the percentage change in…
If the kinetic energy of a particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is
75
25
50
5
Solution
Let the initial de Broglie wavelength be \(\lambda_0=\frac{\mathrm{h}}{\mathrm{mv}_{0}}\)
The kinetic energy is increased 16 times.
Thus \(\mathrm{KE}^{\prime}=16 \mathrm{KE}_0\)
\(\begin{aligned}
& \Rightarrow \frac{1}{2} \mathrm{mv}^2=16 \times \frac{1}{2} \mathrm{mv}_{0}^2 \\
& \Rightarrow \mathrm{v}=4 \mathrm{v}_{0}
\end{aligned}\)
Thus the new de Broglie wavelength \(=\lambda=\frac{h}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{m}\left(4 \mathrm{v}_{0}\right)}=\frac{\lambda_0}{4}\)
Thus the percentage change in wavelength:
\(\begin{aligned}
& =\frac{\lambda_0-\lambda}{\lambda_0} \times 100 \% \\
& =75
\end{aligned}\)