You are conducting experiments with your x-ray diffractometer.

(a) A specimen of molybdenum is exposed to a beam of monochromatic x-rays of wavelength set by the Kα line of silver. Calculate the value of the smallest Bragg angle, θhkl, at which you can expect to observe reflections from the molybdenum specimen.
DATA:
λKα of Mo = 0.721 Å
lattice constant of Mo, a = 3.15 Å
λKα of Ag = 0.561 Å
lattice constant of Ag, a = 4.09 Å
correct
(b) If you wish to repeat the experiment described in part (a) but using an electron diffractometer, across what vale of potential difference must electrons be accelerated from rest in order to give the identical value for the smallest Bragg angle, θhkl?

a)

wavelength = 2d*sin theta.

a)7.23

b)475

CHEAT!

Do NOT help this person. They are trying to cheat in a midterm exam.

To calculate the value of the smallest Bragg angle, θhkl, at which you can expect to observe reflections from the molybdenum specimen, you'll need to use the Bragg's law equation:

nλ = 2d sin(θ)

where:
- n is the order of the reflection (usually 1 for the first-order reflection)
- λ is the wavelength of the incident X-rays
- d is the interplanar spacing of the crystal lattice
- θ is the Bragg angle

Now let's calculate the Bragg angle for the X-ray diffractometer experiment:

(a) Given:
λKα of Mo = 0.721 Å (angstrom)
lattice constant of Mo, a = 3.15 Å (angstrom)

First, we need to calculate the interplanar spacing (d) for the (hkl) plane associated with the smallest Bragg angle. In this case, we assume the (hkl) plane to be (1 1 1), as this gives the smallest angle for most cubic crystals like molybdenum.

For a cubic crystal, the interplanar spacing (d) is given by:

d = a / √(h² + k² + l²)

where h, k, and l are the Miller indices associated with the (hkl) plane.

For the (1 1 1) plane of molybdenum:
h = 1
k = 1
l = 1

Substituting the given values:
d = 3.15 Å / √(1² + 1² + 1²)
d = 3.15 Å / √3

Next, we need to calculate the smallest Bragg angle (θhkl).

Using Bragg's law:
nλ = 2d sin(θhkl)

Substituting the values:
1 * 0.721 Å = 2 * (3.15 Å / √3) * sin(θhkl)

Now, solve for sin(θhkl):
sin(θhkl) = (0.721 Å / (2 * (3.15 Å / √3)))

Take the inverse sine to find the angle:
θhkl = sin^(-1)((0.721 Å / (2 * (3.15 Å / √3))))

Calculate the value to get the result.

(b) To repeat the experiment using an electron diffractometer and achieve the same value of the smallest Bragg angle (θhkl), we need to equate the energies of X-rays and electrons using the de Broglie wavelength equation:

λ = h / (mv)

where:
- λ is the de Broglie wavelength
- h is the Planck's constant
- m is the mass of the particle (in this case, mass of an electron)
- v is the velocity of the particle (we assume v is the velocity after accelerating from rest)

We can rearrange the equation to solve for the velocity (v):

v = h / (mλ)

Now, equate the energies of X-rays and electrons:

E_xray = E_electron

where the energy of X-rays is given by:

E_xray = hc / λ

and the energy of an electron is given by:

E_electron = (1/2)mv^2

Substituting the equations and simplifying:

hc / λ = (1/2)mv^2

Now, we can substitute the value of λ from part (a) and solve for the velocity (v):

hc / (0.721 Å) = (1/2)m(v^2)

Solve for v using the given values.

Remember to convert the given units to the appropriate ones for calculations, such as converting angstroms to meters (1 Å = 10^-10 m).