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Question

(D) The de Broglie wavelength is given by the formula:$\lambda = \frac{h}{\sqrt{2mK}}$ ...... $(i)$where $m$ is the mass and $K$ is the kinetic energy

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(D) The de Broglie wavelength is given by the formula:$\lambda = \frac{h}{\sqrt{2mK}}$ ...... $(i)$where $m$ is the mass and $K$ is the kinetic energy of the particle.When the kinetic energy of the particle is increased to $16$ times its initial value,the new kinetic energy $K' = 16K$.The new de Broglie wavelength $\lambda'$ becomes:$\lambda' = \frac{h}{\sqrt{2m(16K)}} = \frac{h}{4\sqrt{2mK}} = \frac{\lambda}{4}$ (Using equation $(i)$).The percentage change in the de Broglie wavelength is calculated as:$	ext{Percentage change} = \frac{\lambda - \lambda'}{\lambda} 	imes 100$$= \left(1 - \frac{\lambda'}{\lambda}ight) 	imes 100$$= \left(1 - \frac{1}{4}ight) 	imes 100 = \frac{3}{4} 	imes 100 = 75\%$.

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

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