A hydrogen atom transitions from n= 5 state to the ground state

[ikihi]

Homework Statement

A hydrogen atom transitions from n_i= 5 state down to the ground state.
a) What is the energy of photon emitted from the transition of the hydrogen atom?
b) What is the ratio of the momentum of the emitted photon to the momentum of an electron which possesses the same kinetic energy as that of the photon. That is calculate pe⁻/p_γ.

Homework Equations

ΔE_photon = E_f - E_i

E_n= -13.6 eV / (n)²

The Attempt at a Solution

a)
ΔE_photon = E₁ - E₅
ΔE_photon = -13.6 eV [1/(1)² - 1/(5)²] = - 13.0 eV = - 2.08 x 10 ^-18 J (shouldn't the answer here be positive?)

b)
Momentum of electron
P_electron = √(2 ⋅ m ⋅ E_photon)
P_electron = √(2 ⋅ 9.11 x 10^-31 ⋅ 2.08 x 10 ^-18)
P_electron = 1.95 x 10^-24

Momentum of photon
P_photon = h /λ = E_photon / c = (2.08 x 10^-18) / (3.00 x 10⁸)= 6.97 x 10^-27

Ratio
P_electron / P_photon = 1.95 x 10^-24 / 6.97 x 10^-27 = 280

 

[mfb]

The answer should be positive. Your initial formula gives the correct sign if you absorb a photon.

Working with eV everywhere would make calculations easier, but working with SI units is possible as well, of course.

 

[ikihi]

The answer should be positive. Your initial formula gives the correct sign if you absorb a photon.

Working with eV everywhere would make calculations easier, but working with SI units is possible as well, of course.

So I'm using the wrong formula for photon emission?

 

[mfb]

You can just change the sign at the end. Or use the absolute value of the energy difference, then it works in both cases.
Signs should come from understanding the situation. Good if the formulas also have the correct sign, but you should understand which sign the result has to have.