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mars1129 [50]
3 years ago
15

You are doing x-ray diffraction on a crystal that has a cubic structure, using 0.340-nm x rays. Part A If the lattice spacing is

d = 6.70 × 10−10 m, what is the greatest Bragg angle at which you observe peaks in the intensity of the diffracted rays?
Physics
1 answer:
Mrac [35]3 years ago
6 0

Answer:

∅ = 0.26°

Explanation:

Bragg's law states that:

n×λ = 2×d×sin(∅)

<em>where:</em>

  • <em>d is the lattice spacing.</em>
  • <em>λ is the wavelength of the rays.</em>
  • <em>n is the order of the diffracted rays, n = 1.</em>

then:

sin(∅) = λ/2×d

    ∅ = sin^-1(λ/2×d)

    ∅ = 0.26°

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If you lift two loads up one story, how much work do you do compared to lifting just one load up one story?
Daniel [21]

Answer:a

Explanation:

We have to lift the load two loads up one story, so energy required is

let m be the mass of each load and h is the height of each story

Energy E_1=2\times m\times g\times \h

E_1=2mgh

Here energy gained is the potential energy which depends upon the datum(floor).

For lifting one load up one story

Energy required

E_2=mgh

thus E_2 is half of E_1

So option a is correct                                

6 0
3 years ago
Unpolarized light of intensity Io is incident on a stack of 7 polarizing filters, each with its axis rotated 17°cw with respect
mash [69]

The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925

To answer the question, we need to know what polarization of light is.

<h3>What is polarization of light?</h3>

This is when the electric field vector of light is oscillating in one plane.

  • Now for light of intensity I' which is initially unpolarized, its intensity after polarization is I = 1/2I'.
  • Also, for light initially polarized, its intensity after polarization is I"' = I"cos²Ф where Ф is the angle between the initial direction and the direction of polarization.
<h3>Intensity of light through each polarized filter</h3>

Given that we have 7 polarizing filters, each rotated 17° cw with respect to the previous filter.

So, since the light is initially unpolarized,

  • The intensity through the first polarizing filter is I₁ = 1/2I₀ where I₀ is the initial intensity.
  • The intensity through the second polarizing filter is I₂ = I₁cos²17°= 1/2I₀cos²17°
  • The intensity through the third polarizing filter is I₃ = I₂cos²17° = 1/2I₀cos⁴17°
  • The intensity through the fourth polarizing filter is I₄ = I₃cos²17° = 1/2I₀cos⁶17°
  • The intensity through the fifth polarizing filter is I₅ = I₄cos²17° = 1/2I₀cos⁸17°
  • The intensity through the sixth polarizing filter is I₆ = I₅cos²17° = 1/2I₀cos¹⁰17°
  • The intensity through the seventh polarizing filter is I₇ = I₆cos²17° = 1/2I₀cos¹²17°.
<h3>The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity</h3>

Since I₇ is the last intensity I₇ = It = 1/2I₀cos¹²17°.

So, It/I₀ = 1/2cos¹²17°

= 1/2(0.9563)¹²

= 1/2 × 0.5850

= 0.2925

So, the ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925

Learn more about intensity of polarized light here:

brainly.com/question/25402491

5 0
2 years ago
Two electrons in a vacuum exert force of F = 3.8E-09 N on each other. They are then moved such that they are separated by x = 8.
iren [92.7K]

Answer:

F_n = 5.65E-11 N

d =  1.20682E-31 m

Explanation:

F = 3.8E-09 N

where

m = Mass of electron = 9.109E−31 kilograms

G = Gravitational constant = 6.67E-11 m³/kgs²

x = Distance between them

F=G\frac{m^2}{x^2}\\\Rightarrow 3.8E-09=G\frac{m^2}{x^2}

For F_n

F_n=G\frac{m^2}{x^2}\\\Rightarrow F_n=G\frac{m^2}{(8.2x)^2}\\\Rightarrow F_n=G\frac{m^2}{67.24x^2}

Dividing the above equations we get

\frac{F}{F_n}=\frac{G\frac{m^2}{x^2}}{G\frac{m^2}{67.24x^2}}\\\Rightarrow \frac{F}{F_n}=67.24\\\Rightarrow F_n=\frac{F}{67.24}\\\Rightarrow F_n=\frac{3.8E-09}{67.24}\\\Rightarrow F_n=5.65E-11\ N

F_n = 5.65E-11 N

F=G\frac{m^2}{x^2}\\\Rightarrow x=\sqrt{\frac{Gm^2}{F}}\\\Rightarrow x=\sqrt{\frac{G}{F}}m\\\Rightarrow x=\sqrt{\frac{6.67E-11}{3.8E-09}}9.109E-31\\\Rightarrow x=1.20682E-31\ m

d =  1.20682E-31 m

8 0
4 years ago
the force shown in figure 7-15 moves an object from x=0 to x=0.75 m. How much work is done by the force?
Phantasy [73]
Work is force multiplied by the distance the force moves the object
6 0
3 years ago
1. Predict whether the energy required to remove an electron from magnesium and potassium would be more or less than that requir
Inga [223]

In both cases less energy is required

But comparetively Mg require more energy than K

Let's see the electron configuration of Both

  • [Mg]=1s²2s²2p⁶3s²=[Ne]3s²
  • [K]=1s²2s²2p⁶3s²3p⁶4s¹=[Ar]4s¹

K has only one valence electron so very less ionization enthalpy so less energy required

Mg has 2 so more IE hence more energy required

8 0
2 years ago
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