The de-Broglie wavelength of an electron travelling with $10 %$ of velocity of light is equal to:

242.4 p m

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24.2 p m

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2.42 p m

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0.2424 p m

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Solution

The correct option is B $24.2 p m$
Given, Velocity of light $= 0.1$ $C = 0.1 (3 \times 10^{8}) = 3 \times 10^{7}$ $m / s$
de- Broglie wavelength $= \frac{h}{m v} = \frac{6.626 \times 10^{- 34}}{9.1 \times 10^{- 31} \times 3 \times 10^{7}} = 0.242 \times 10^{- 10}$ $m = 24.2 \times 10^{- 12}$ $m = 24.2$ $p m$

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The de-Broglie wavelength of an electron travelling with 10% of velocity of light is equal to:

The correct option is B 24.2pmGiven, Velocity of light =0.1 C=0.1(3×108)=3×107 m/sde- Broglie wavelength =hmv=6.626×10−349.1×10−31×3×107=0.242×10−10 m=24.2×10−12 m=24.2 pm

The de-Broglie wavelength of an electron travelling with $10 %$ of velocity of light is equal to: