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
is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)