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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Initially, let the velocity be $v$, so $\lambda_1 = \frac{h}{mv}$.When the velocity

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$100\%$ increase

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Initially, let the velocity be $v$, so $\lambda_1 = \frac{h}{mv}$.When the velocity is reduced to half, the new velocity is $v' = \frac{v}{2}$.The new wavelength is $\lambda_2 = \frac{h}{m(v/2)} = 2 \left( \frac{h}{mv} ight) = 2\lambda_1$.The change in wavelength is $\Delta \lambda = \lambda_2 - \lambda_1 = 2\lambda_1 - \lambda_1 = \lambda_1$.The percentage change is $\frac{\Delta \lambda}{\lambda_1} 	imes 100 = \frac{\lambda_1}{\lambda_1} 	imes 100 = 100\%$.Since the wavelength increases, it is a $100\%$ increase.

If the velocity of a moving particle is reduced to half, what will be the percentage change in its de Broglie wavelength?

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