A photon collides with an electron and gets scattered through an angle of $90^{0}$ . The electron recoils and moves in another direction. The compton wavelength is $(h = 6.62 \times 10^{- 34} J s .)$

0.121 \times 10^{- 11} m

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0.486 \times 10^{- 11} m

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2.4 \times 10^{- 11} m

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0.243 \times 10^{- 11} m

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Solution

The correct option is C $0.243 \times 10^{- 11} m$
Compton wavalength $= \frac{h}{m_{e} c} (1 - c o s θ)$

$= \frac{6.62 \times 10^{- 34}}{9.1 \times 10^{- 31} \times 3 \times 10^{8}} (1 - c o s 90^{0})$

$= 0.243 \times 10^{- 11} (1 - 0) (∵ c o s 90^{0} = 0)$
$= 0.243 \times 10^{- 11} m$
So, the answer is option (D).

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Similar questions

Radius of H-atom in its ground state is $5.03 \times 10^{- 11} m .$ After collision with an electron it is found to have a radius of $20.12 \times 10^{- 11} m .$ The principal quantum number $^{'} n^{'}$ of the final state of the atom is:

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A photon collides with an electron and gets scattered through an angle of 900. The electron recoils and moves in another direction. The compton wavelength is (h=6.62×10−34Js.)

The correct option is C 0.243×10−11mCompton wavalength =hmec(1−cosθ) =6.62×10−349.1×10−31×3×108(1−cos 900) =0.243×10−11(1−0) (∵cos 900=0) =0.243×10−11mSo, the answer is option (D).

A photon collides with an electron and gets scattered through an angle of $90^{0}$ . The electron recoils and moves in another direction. The compton wavelength is $(h = 6.62 \times 10^{- 34} J s .)$