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Question: The de Broglie wavelength of an electron with kinetic energy 120 eV is: (Given h= 6.63×10<sup>-34</s...

The de Broglie wavelength of an electron with kinetic energy 120 eV is: (Given h= 6.63×10-34 Js, me =9×10-34 Kg, 1 eV = 1.6×10-19 J)

A

2.13Å

B

1.13Å

C

4.15Å

D

3.14Å

Answer

1.13Å

Explanation

Solution

: As p=2mKp = \sqrt{2mK}

}{= 5.88 \times 10^{- 24}kgms^{- 1}}$$ de Broglie wavelength, $$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{- 34}}{5.88 \times 10^{- 24}}$$ $$= 1.13 \times 10^{- 10}m = 1.13Å$$