Thank you for reporting, we will resolve it shortly
Q.
If the kinetic energy of the particle is increased to $16$ times its previous value, the percentage change in the de Broglie wavelength of the particle is
de Broglie wavelength,
$\lambda=\frac{h}{\sqrt{2 m K}}$
where $m$ is the mass and $K$ is the kinetic energy of the particle.
When kinetic energy of the particle is increased to 16 times, then its de Broglie wavelength becomes,
$\lambda^{\prime}=\frac{h}{\sqrt{2 m(16 K)}}=\frac{1}{4} \frac{\lambda}{\sqrt{2 m K}}=\frac{\lambda}{4}$( Using ( i ))
$\%$ change in the de Broglie wavelength
$=\frac{\lambda-\lambda^{\prime}}{\lambda} \times 100=\left(1-\frac{\lambda^{\prime}}{\lambda}\right) \times 100$
$=\left(1-\frac{1}{4}\right) \times 100=75 \%$