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Question: Calculate de-Broglie wavelength of an electron travelling at 1% of the speed of light...

Calculate de-Broglie wavelength of an electron travelling at 1% of the speed of light

A

2.73×10−242.73 \times 10^{- 24}

B

2.42×10−102.42 \times 10^{- 10}

C

242.2×1010242.2 \times 10^{10}

D

None of these

Answer

2.42×10−102.42 \times 10^{- 10}

Explanation

Solution

One percent of the speed of light is

v=(1100)(3.00×108ms−1)\mathbf{v =}\left( \frac{\mathbf{1}}{\mathbf{100}} \right)\mathbf{(3.00 \times 1}\mathbf{0}^{\mathbf{8}}\mathbf{m}\mathbf{s}^{\mathbf{- 1}}\mathbf{)} = 3.00×106ms−1\mathbf{3.00}\mathbf{\times}\mathbf{1}\mathbf{0}^{\mathbf{6}}\mathbf{m}\mathbf{s}^{\mathbf{-}\mathbf{1}}

Momentum of the electron (p)(p) = mνm\nu

= (9.11×10−31kg)(3.00×106ms−1)(9.11 \times 10^{- 31}kg)(3.00 \times 10^{6}ms^{- 1})

= 2.73×10−24kgms−12.73 \times 10^{- 24}kgms^{- 1}

The de-broglie wavelength of this electron is

λ=hp=6.626×10−342.73×10−24kgms−1\lambda = \frac{h}{p} = \frac{6.626 \times 10^{- 34}}{2.73 \times 10^{- 24}kgms^{- 1}}

λ=2.424×10−10m\lambda = 2.424 \times 10^{- 10}m.