Answer:
Wavelength = 2.91 x 10⁻¹² m, Energy = 6.8 x 10⁻¹⁴
Explanation:
In order to show that a free electron can’t completely absorb a photon, the equation for relativistic energy and momentum will be needed, along the equation for the energy and momentum of a photon. The conservation of energy and momentum will also be used.
E = y(u) mc²
Here c is the speed of light in vacuum and y(u) is the Lorentz factor
y(u) = 1/√[1-(u/c)²], where u is the velocity of the particle
The relativistic momentum p of an object of mass m and velocity u is given by
p = y(u)mu
Here y(u) being the Lorentz factor
The energy E of a photon of wavelength λ is
E = hc/λ, where h is the Planck’s constant 6.6 x 10⁻³⁴ J.s and c being the speed of light in vacuum 3 x 108m/s
The momentum p of a photon of wavelenght λ is,
P = h/λ
If the electron is moving, it will start the interaction with some momentum and energy already. Momentum of the electron and photon in the initial and final state is
p(pi) + p(ei) = p(pf) + p(ef), equation 1, where p refers to momentum and the e and p in the brackets refer to proton and electron respectively
The momentum of the photon in the initial state is,
p(pi) = h/λ(i)
The momentum of the electron in the initial state is,
p(ei) = y(i)mu(i)
The momentum of the electron in the final state is
p(ef) = y(f)mu(f)
Since the electron starts off going in the negative direction, that momentum will be negative, along with the photon’s momentum after the collision
Rearranging the equation 1 , we get
p(pi) – p(ei) = -p(pf) +p(ef)
Substitute h/λ(i) for p(pi) , h/λ(f) for p(pf) , y(i)mu(i) for p(ei), y(f)mu(f) for p(ef) in the equation 1 and solve
h/λ(i) – y(i)mu(i) = -h/λ(f) – y(f)mu(f), equation 2
Next write out the energy conservation equation and expand it
E(pi) + E(ei) = E(pf) + E(ei)
Kinetic energy of the electron and photon in the initial state is
E(p) + E(ei) = E(ef), equation 3
The energy of the electron in the initial state is
E(pi) = hc/λ(i)
The energy of the electron in the final state is
E(pf) = hc/λ(f)
Energy of the photon in the initial state is
E(ei) = y(i)mc2, where y(i) is the frequency of the photon int the initial state
Energy of the electron in the final state is
E(ef) = y(f)mc2
Substitute hc/λ(i) for E(pi), hc/λ(f) for E(pf), y(i)mc² for E(ei) and y(f)mc² for E(ef) in equation 3
Hc/λ(i) + y(i)mc² = hc/λ(f) + y(f)mc², equation 4
Solve the equation for h/λ(f)
h/λ(i) + y(i)mc = h/λ(f) + y(f)mc
h/λ(f) = h/lmda(i) + (y(i) – y(f)c)m
Substitute h/λ(i) + (y(i) – y(f)c)m for h/λ(f) in equation 2 and solve
h/λ(i) -y(i)mu(i) = -h/λ(f) + y(f)mu(f)
h/λ(i) -y(i)mu(i) = -h/λ(i) + (y(f) – y(i))mc + y(f)mu(f)
Rearrange to get all λ(i) terms on one side, we get
2h/λ(i) = m[y(i)u(i) +y(f)u(f) + (y(f) – y(i)c)]
λ(i) = 2h/[m{y(i)u(i) + y(f)u(f) + (y(f) – y(i))c}]
λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]
Calculate the Lorentz factor using u(i) = 0.8c for y(i) and u(i) = 0.6c for y(f)
y(i) = 1/[√[1 – (0.8c/c)²] = 5/3
y(f) = 1/√[1 – (0.6c/c)²] = 1.25
Substitute 6.63 x 10⁻³⁴ J.s for h, 0.511eV/c2 = 9.11 x 10⁻³¹ kg for m, 5/3 for y(i), 0.8c for u(i), 1.25 for y(f), 0.6c for u(f), and 3 x 10⁸ m/s for c in the equation derived for λ(i)
λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]
λ(i) = 2(6.63 x 10-34)/[(9.11 x 10-31)(3 x 108){(5/3)(0.8) + (1.25)(0.6) + ((1.25) – (5/3))}]
λ(i) = 2.91 x 10⁻¹² m
So, the initial wavelength of the photon was 2.91 x 10-12 m
Energy of the incoming photon is
E(pi) = hc/λ(i)
E(pi) = (6.63 x 10⁻³⁴)(3 x 10⁸)/(2.911 x 10⁻¹²) = 6.833 x 10⁻¹⁴ = 6.8 x 10⁻¹⁴
So the energy of the photon is 6.8 x 10⁻¹⁴ J
(1)
, 
and 
, so the final pressure of the gas can be found by re-arranging eq.(1):