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kondaur [170]
3 years ago
5

If the planes of a crystal are 3.50 (1 A= 10^-10m = Ångstrom unit) apart, what wavelength of electromagnetic waves are needed so

that the first strong interference maximum in the Bragg reflection occurs when the waves strike the planes at an angle of 22.0 degrees?
Physics
1 answer:
goldenfox [79]3 years ago
4 0

Answer:

λ = 2.62 x 10⁻¹⁰ m = 0.262 nm

Explanation:

We can use Bragg's Law's equation to solve this problem. The Bragg's Law's equation is written as follows:

mλ = 2d Sin θ

where,

m = order of reflection = 1

λ = wavelength = ?

d = distance between the planes of crystal = 3.5 x 10⁻¹⁰ m

θ = strike angle of waves on plane = 22°

Therefore, substituting the respective values in the equation, we get:

(1)λ = (2)(3.5 x 10⁻¹⁰ m)(Sin 22°)

<u>λ = 2.62 x 10⁻¹⁰ m = 0.262 nm</u>

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The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
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This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

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FA = NB = W / 2tan∅

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Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

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FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

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