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sergiy2304 [10]

3 years ago

6

An x-ray beam with wavelength 0.240 nm is directed at a crystal. as the angle of incidence increases, you observe the first stro

ng interference maximum at an angle 24.5 ∘. what is the spacing d between the planes of the crystal?

1 answer:

jeka943 years ago

5 0

This phenomenon is called the Bragg scattering. The working equation used for this is:

λ = 2dsinθ
where
λ is the wavelength and θ is the incident angle

0.240nm = 2dsin(24.5°)
Solving for d,
<em>d = 0.29 nm</em>

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Given the description of period and amplitude, the SHM could be described by:

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$f(x)=5\, \,sin(\frac{\pi}{4}x)\\f'(x)=5\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$

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In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

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Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

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$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

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