Question

Electrons accelerated by potential V are diffracted from a crystal. If $d=1A_{o}$ and $i=30_{∘}$ , V should be about :

$(h=6.6×10_{−34}js,m=9.1×10_{−31}kg,e=1.6×10_{−19}c)$

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Correct option is

$2dsinθ=nλ$

angle of incidence = $90−i$

Given $i=30_{o}$, $d=1×10_{−10}m$, $h=6.6×10_{−34}$, $m=9.1×10_{−31}$ and $e=1.6×10_{−19}$

Also $n=1$ for $1_{st}$ order diffraction.

Therefore,

$2×1×10_{−10}sin(90−30)=1×λ$

$⟹λ=2×10_{−10}23 =3 ×10_{−10}m$

We have de-Broglie wavelength, $λ=mvh $

and also $21 mv_{2}=eV$ $⟹v=m2eV $

Substituting in de-Broglie wavength, we get,

$λ=2meV h $= $3 ×10_{−10}$m

$⟹V=49.86V$

Solve any question of Dual Nature of Radiation And Matter with:-

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