In Davisson-Germer experiment, the wavelength associated with nickel crystal is

Question

In Davisson-Germer experiment, the wavelength associated with nickel crystal is (A) 1.66 mathringA (B) 2 mathringA (C) 2.3 mathringA (D) 3.86 mathringA

Tardigrade Admin · Accepted Answer

For the nickel crystal, the interatomic separation is

d = 0.91 \overset{˚}{A}

According to Bragg's law, for first order diffraction maxima

(n = 1)

, we have

2 d sin θ = 1 \times λ

∴ λ = 2 \times 0.91 \times sin 6 5^{\circ} = 1.65 \overset{˚}{A}

Q. In Davisson-Germer experiment, the wavelength associated with nickel crystal is

$1.66 \overset{˚}{A}$

$2 \overset{˚}{A}$

$2.3 \overset{˚}{A}$

$3.86 \overset{˚}{A}$

For the nickel crystal, the interatomic separation is
$d = 0.91 \overset{˚}{A}$
According to Bragg's law, for first order diffraction maxima $(n = 1)$ , we have
$2 d sin θ = 1 \times λ$
$∴ λ = 2 \times 0.91 \times sin 6 5^{\circ} = 1.65 \overset{˚}{A}$

Q. In Davisson-Germer experiment, the wavelength associated with nickel crystal is

1.66A˚

2A˚

2.3A˚

3.86A˚

For the nickel crystal, the interatomic separation is d=0.91A˚ According to Bragg's law, for first order diffraction maxima (n=1), we have 2dsinθ=1×λ ∴λ=2×0.91×sin65∘=1.65A˚

$1.66 \overset{˚}{A}$

$2 \overset{˚}{A}$

$2.3 \overset{˚}{A}$

$3.86 \overset{˚}{A}$

For the nickel crystal, the interatomic separation is
$d = 0.91 \overset{˚}{A}$
According to Bragg's law, for first order diffraction maxima $(n = 1)$ , we have
$2 d sin θ = 1 \times λ$
$∴ λ = 2 \times 0.91 \times sin 6 5^{\circ} = 1.65 \overset{˚}{A}$