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Question: An electron has kinetic energy\(2.8 \times 10^{- 23}J\). de-Broglie wavelength will be nearly \((m_{...

An electron has kinetic energy2.8×10−23J2.8 \times 10^{- 23}J. de-Broglie wavelength will be nearly (me=9.1×10−31kg)(m_{e} = 9.1 \times 10^{- 31}kg)

A

9.28×10−4m9.28 \times 10^{- 4}m

B

9.28×10−7m9.28 \times 10^{- 7}m

C

9.28×10−8m9.28 \times 10^{- 8}m

D

9.28×10−10m9.28 \times 10^{- 10}m

Answer

9.28×10−8m9.28 \times 10^{- 8}m

Explanation

Solution

Formula for de-Broglie wavelength is

λ=hp\lambda = \frac{h}{p} or λ=hmv⇒eV=12mv2\lambda = \frac{h}{mv} \Rightarrow eV = \frac{1}{2}mv^{2} or ν=2eVm\nu = \sqrt{\frac{2eV}{m}}

λ=h2meV=6.62×10−342×9.1×10−31×2.8×10−23\lambda = \frac{h}{\sqrt{2meV}} = \frac{6.62 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 2.8 \times 10^{- 23}}}

λ=9.28×10−8meter\lambda = 9.28 \times 10^{- 8}meter.