Chapter 4: Q23E (page 135)

Question: Atoms in a crystal form atomic planes at many different angles with respect to the surface. The accompanying figure shows the behaviors of representative incident and scattered waves in the Davisson-Germer experiment. A beam of electrons accelerated through 54 V is directed normally at a nickel surface, and strong reflection is detected only at an angle $ϕ$ of 50⁰.Using the Bragg law, show that this implies a spacing D of nickel atoms on the surface in agreement with the known value of 0.22 nm.

Short Answer

Expert verified

Answer:

It is shown that D = 0.22nm .

Step by step solution

de Broglie’s formula

The de Broglie’s wavelength formula is $λ = \frac{h}{m v}$ .

Proof

Using the formula , $V = \frac{h^{2}}{2 m q λ^{2}}$ find the value of as follows:

$54 V = \frac{{(6.63 \times 10^{- 34} J \cdot s)}^{2}}{2 (9.11 \times 10^{- 31} k g) (1.6 \times 10^{- 19} C) λ^{2}} λ^{2} = \frac{{(6.63 \times 10^{- 34} J \cdot s)}^{2}}{108 V (9.11 \times 10^{- 31} k g) (1.6 \times 10^{- 19} C)} λ = \frac{6.63 \times 10^{- 34} J \cdot s}{\sqrt{108 V (9.11 \times 10^{- 31} k g) (1.6 \times 10^{- 19} C)}} λ = 1.67 \times 10^{- 10} m$

From the figure, it is clear that the angle between the incident beam and atomic plate can be obtained as follows:

$θ = 90 ° - \frac{ϕ}{2} = 90 ° - \frac{50 °}{2} = 90 ° - 25 ° = 65 °$

Using Bragg’s law, we have:

$2 d \sin θ = m λ d = \frac{m λ}{2 \sin θ} d = \frac{1 \times 1.67 \times 10^{- 10} m}{2 \times \sin 65 °} d = 0.092 nm$

The angle between the planes and the crystal surface can be obtained as follows:

$θ = 90 ° - 65 ° = 25 °$

Now, by the sine law, we get:

$\sin θ = \frac{d}{D} D = \frac{d}{\sin θ} D = \frac{0.092 n m}{\sin 25 °} D = 0.22 nm$

Thus, it is proved that spacing has the value D of 0.22nm.

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de Broglie’s formula

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