There are four particles A, B, C and D with the same mass and equal kinetic energy K. If the kinetic energy of particle A becomes 4 K, B becomes 9 K, C becomes 64 K and D becomes 16 K times, respectively, the fractional change in the de Broglie wavelength of the particles arranged in increasing order is

(C), (D), (B), (A)

The correct answer is Option (1) → (A), (B), (D), (C)

$\lambda \propto \frac{1}{\sqrt{K}}$

$\text{If }K \to mK\text{ then } \frac{\lambda_{\text{new}}}{\lambda_{\text{old}}}=\sqrt{\frac{1}{m}}$

$\text{Fractional change }=\frac{\lambda_{\text{new}}-\lambda_{\text{old}}}{\lambda_{\text{old}}}=\sqrt{\frac{1}{m}}-1$

For A: $m=4\Rightarrow \sqrt{\frac{1}{4}}-1=\frac{1}{2}-1=-\frac{1}{2}$

For B: $m=9\Rightarrow \sqrt{\frac{1}{9}}-1=\frac{1}{3}-1=-\frac{2}{3}$

For C: $m=64\Rightarrow \sqrt{\frac{1}{64}}-1=\frac{1}{8}-1=-\frac{7}{8}$

For D: $m=16\Rightarrow \sqrt{\frac{1}{16}}-1=\frac{1}{4}-1=-\frac{3}{4}$

$\text{Arrange in increasing order (most negative → least negative): }C\;(-\frac{7}{8})\;,\;D\;(-\frac{3}{4})\;,\;B\;(-\frac{2}{3})\;,\;A\;(-\frac{1}{2})$

${C\;<\;D\;<\;B\;<\;A}$