If a diffraction peak is observed when electrons are incident at an angle $^{'} i^{'}$ from the normal to the crystal planes with distance $^{'} d^{'}$ between them (see figure), de-Broglie wavelength $λ_{d B}$ of electrons can be calculated by the relationship ( $n$ is an integer):

d sin i = n λ_{d B}

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2 d cos i = n λ_{d B}

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2 d sin i = n λ_{d B}

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d cos i = n λ_{d B}

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Solution

The correct option is B $2 d cos i = n λ_{d B}$
From condition of constructive interference of Bragg's law,
$2 d s i n θ = n λ_{d b}$
but here $θ = 90 - i$
So,
$2 d s i n (90 - i) = n λ_{d b}$
$\Rightarrow 2 d c o s i = n λ_{d b}$

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If a diffraction peak is observed when electrons are incident at an angle ′i′ from the normal to the crystal planes with distance ′d′ between them (see figure), de-Broglie wavelength λdB of electrons can be calculated by the relationship (n is an integer):

The correct option is B 2dcosi=nλdBFrom condition of constructive interference of Bragg's law,2dsinθ=nλdbbut here θ=90−iSo,2dsin(90−i)=nλdb⇒2dcosi=nλdb

If a diffraction peak is observed when electrons are incident at an angle $^{'} i^{'}$ from the normal to the crystal planes with distance $^{'} d^{'}$ between them (see figure), de-Broglie wavelength $λ_{d B}$ of electrons can be calculated by the relationship ( $n$ is an integer):

The correct option is B $2 d cos i = n λ_{d B}$
From condition of constructive interference of Bragg's law,
$2 d s i n θ = n λ_{d b}$
but here $θ = 90 - i$
So,
$2 d s i n (90 - i) = n λ_{d b}$
$\Rightarrow 2 d c o s i = n λ_{d b}$