Question

If kinetic energy of a moving particle is increased by 300%, then de-Broglie wavelength associated with it.

Answer 1

De - Broglie's wavelength equation in term of kinetic energy is given by,

$\boxed{\bf{\lambda=\frac{h}{\sqrt{2Km}}}}$

where h is plank's constant, m is mass of particle and K is kinetic energy of particle.

here it is clear that De-broglie wavelength is inversely proportional to square root of kinetic energy .

e.g., $\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{K_2}{K_1}}$

if we assume initial kinetic energy of particle is K and wavelength due to it is $\lambda$

final kinetic energy = K + 300% of K = 4K

now, $\frac{\lambda}{\lambda_2}=\sqrt{\frac{4K}{K}}=2$

or, $\lambda_2=\frac{\lambda}{2}$

hence, final wavelength is half of initial wavelength. hence, it is decreased by 50% .