Question 273: Dual Nature of Matter and Radiation

Question

The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of ' $n$ ' is

Accepted Answer

Step 1: Relation between momentum and wavelength
DeÃ¢ÂÂBroglie wavelength:
$ \lambda = \frac{h}{p}. $
If wavelength reduces from $\lambda$ to $	frac{\lambda}{2}$:
$ \lambda' = \frac{\lambda}{2} \quad \Rightarrow \quad p' = \frac{h}{\lambda'} = \frac{h}{\lambda/2} = \frac{2h}{\lambda} = 2p. $
So new momentum = $2p$.
Step 2: Kinetic energy in terms of momentum
For nonÃ¢ÂÂrelativistic electron:
$ E = \frac{p^2}{2m}. $
Initial energy:
$ E_i = \frac{p^2}{2m}. $
Final energy:
$ E_f = \frac{p'^2}{2m} = \frac{(2p)^2}{2m} = \frac{4p^2}{2m} = 4E_i. $
Step 3: Energy that should be added
Energy added:
$ \Delta E = E_f - E_i = 4E_i - E_i = 3E_i. $
So the added energy is 3 times the initial energy.
Final Answer:
$ n = 3 $
Correct Option: B (3).
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4

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3

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2

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1

Dual Nature of Matter and Radiation

The energy that should be added to an electron to reduce its de-Broglie wavelength from \lambda to \frac{\lambda}{2} is n times the initial energy. The value of ' n ' is

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