An electron in the hydrogen atom makes a transition from an energy level of -1.51 eV to one of energy -3.40 eV.In this process a photon is emitted with what wavelength?
A)1.89 nm
B)656 nm
C)456 nm
D)821 nm
E)365 nm

Question

Quizplus · Accepted Answer

We can use the formula for the energy of a photon: 
E = hc/λ 
where E is the energy of the photon, h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon. 
The energy difference between the two energy levels of the electron is: 
ΔE = -3.40 eV - (-1.51 eV) = -1.89 eV 
We need to convert this energy difference to joules, so we can use the formula: 
1 eV = 1.602 x 10^-19 J 
ΔE = (-1.89 eV) x (1.602 x 10^-19 J/eV) = -3.03 x 10^-19 J 
Now we can plug this into the formula for the energy of the photon: 
-3.03 x 10^-19 J = (6.626 x 10^-34 J.s)(3.00 x 10^8 m/s)/λ 
Solving for λ gives: 
λ = 656 nm

An Electron in the Hydrogen Atom Makes a Transition from an Energy