\(\lambda=\frac{h}{\sqrt{2 m K}}\)           ...... \((i)\)

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,

\({\lambda ^\prime } = \frac{h}{{\sqrt {2m(16K)} }}\) \( = \frac{1}{4}\frac{\lambda }{{\sqrt {2mK} }} = \frac{\lambda }{4}\) (Using  \((i)\) )

\(\%\) change in the de Broglie wavelength

\({=\frac{\lambda-\lambda^{\prime}}{\lambda} \times 100=\left(1-\frac{\lambda^{\prime}}{\lambda}\right) \times 100}\)

\( = \left( {1 - \frac{1}{4}} \right) \times 100 = 75\% \)