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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:
A
25
B
75
C
60
D
50
Detailed Solution
$\lambda = \frac{h}{\sqrt{2mK}}$
$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{K_2}{K_1}} = \sqrt{16} = 4 \Rightarrow \lambda_2 = \frac{\lambda_1}{4}$
% change $= \frac{1 - 4}{4}\times100 = -75\%$, i.e. a decrease of 75%.
$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{K_2}{K_1}} = \sqrt{16} = 4 \Rightarrow \lambda_2 = \frac{\lambda_1}{4}$
% change $= \frac{1 - 4}{4}\times100 = -75\%$, i.e. a decrease of 75%.