A particle of mass ${6.6×10}^{-30} kg$ 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 $10^{-4} m/s$ $({Take h=6.6×10}^{-34} J-s)$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

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

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

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

answer is A, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$λ = \frac{h}{m v} \frac{d x}{d t} = \frac{h}{m v^{2}} (\frac{d v}{d t}) = 10^{- 4} - \int_{10}^{2} \frac{d v}{v^{2}} = \frac{10^{- 4} m}{h} \int_{0}^{0.9} d t (\frac{1}{V} - \frac{1}{10}) = \frac{10^{- 4} \times 6.6 \times 10^{- 30}}{6.6 \times 10^{- 34}} (0.9) \frac{1}{V} - 0.1 = 0.9 V = 1 m / s a = + \frac{m V^{2}}{h} (\frac{d λ}{d t}) = \frac{6.6 \times 10^{- 30}}{6.6 \times 10^{- 34}} \times {(1)}^{2} \times 10^{- 4} = 1 m / s^{2}$