Atomic StructureHard

Question

An electron is continuously accelerated in a vacuum tube by the application of a potential di­fference. If its de Broglie wavelength decreases by 1% over a path length of l cm, then its kinetic energy

Options

A.increases by 1%
B.increases by 2%
C.decreases by 2%
D.increases by 0.5%

Solution

$\lambda = \frac{h}{\sqrt{2mE}} \Rightarrow \frac{E_{2}}{E_{1}} = \left( \frac{\lambda_{1}}{\lambda_{2}} \right)^{2} = \left( \frac{100}{99} \right)^{2} = 1.02$

∴ E2 is about 2% greater than E1.

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