A particle of mass $6.6 \times 10^{- 30} k g$ starts (t = 0) moving on a straight line with velocity 10 m/s. Its velocity decreases with time, however rate of change of de-Broglie wavelength associated with particle remains constant at $10^{- 4}$ m/s. (Take $h = 6.6 \times 10^{- 34} J - s$ ) :-

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Velocity of particle at t = 0.9s is 1 m/s

Velocity of particle at t = 0.9s is 6 m/s

Magnitude of retardation of particle at t = 0.9 s is 4 $m / s^{2}$

Magnitude of retardation of particle at t = 0.9 s is 1 $m / s^{2}$

answer is A, D.

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Detailed Solution

$λ = \frac{h}{m v} \frac{d h}{d t} = \frac{h}{m v^{2}} \frac{d v}{d t} 10^{- 4} = - \frac{10^{- 34} \times 6.6}{6.6 \times 10^{- 30}} \frac{1}{v^{2}} \frac{d v}{d t} \frac{d v}{d t} = - v^{2} \int_{10}^{v} \frac{d v}{v^{2}} = - \int_{0}^{t} d t \frac{1}{v} - \frac{1}{10} = t \frac{1}{v} = \frac{1 + 10 t}{10} v = \frac{10}{1 + 10 t}$