JEE PYQ: Atomic Physics Question 44
Question 44 - 2020 (09 Jan Shift 1)
A particle moving with kinetic energy E has de Broglie wavelength $\lambda$. If energy $\Delta E$ is added to its energy, the wavelength becomes $\frac{\lambda}{2}$. Value of $\Delta E$ is:
(1) E
(2) 4E
(3) 3E
(4) 2E
Show Answer
Answer: (3)
Solution
Using $\lambda = \frac{h}{\sqrt{2mKE}}$. $\frac{\lambda}{2} = \frac{h}{\sqrt{2m(KE + \Delta E)}}$. $\frac{\lambda}{\lambda/2} = \sqrt{\frac{KE + \Delta E}{KE}}$. $4 = \frac{KE + \Delta E}{KE}$. $\Delta E = 3KE = 3E$.
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