A beam of electrons of energy $E$ scatters from a target having atomic spacing of $1$ Å. The first maximum intensity occurs at $\theta = 60^\circ$. Then $E$ (in eV) is ______. (Planck constant $h = 6.63 \times 10^{-34}$ J·s, electron mass $m = 9.1 \times 10^{-31}$ kg)

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Modern Physics

A beam of electrons of energy

E

scatters from a target having atomic spacing of

1

Å. The first maximum intensity occurs at

θ = 6 0^{\circ}

. Then

E

(in eV) is ______.
(Planck constant

h = 6.63 \times 1 0^{- 34}

J·s, electron mass

m = 9.1 \times 1 0^{- 31}

kg)

Answer: 50.47

Solution

The electrons act as waves of de Broglie wavelength

λ = \frac{h}{2 m E}

, and the first diffraction maximum from planes spaced

d

apart obeys

2 d sin θ = λ

. With

d = 1

A and

θ = 6 0^{\circ}

λ = 2 \times 1 0^{- 10} \times sin 6 0^{\circ} = 3 \times 1 0^{- 10}

m. Then

E = \frac{h ^{2}}{2 m λ ^{2}} = \frac{( 6.63 \times 1 0 ^{- 34} ) ^{2}}{2 \times 9.1 \times 1 0 ^{- 31} \times 3 \times 1 0 ^{- 20}} \approx 8.07 \times 1 0^{- 18}

J, and dividing by

1.6 \times 1 0^{- 19}

J/eV gives

E \approx 50.47

eV.

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Numerical answer

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