Question

A ground state hydrogen atom absorbs a photon of light having a wavelength of 92.05 nm. It then gives off a photon having a wavelength of 1736 nm. What is the final state of the hydrogen atom?

Answer 1

Absorbed photon energy
Ea = hc/λ.. (Planck's equation)
Ea = hc / 92.05^-9m 

Energy emitted
Ee = hc/ 1736^-9m 

Energy retained ..
∆E = Ea - Ee = hc(1/92.05^-9 - 1/1736^-9) 
∆E = (6.625^-34)(3.0^8) (1.028^7)
∆E = 2.04^-18 J 

Converting J to eV (1.60^-19 J/eV)
 ∆E = 2.04^-18 / 1.60^-19
∆E = 12.70 eV 

Ground state (n=1) energy for Hydrogen = - 13.60eV 

New energy state = (-13.60 + 12.70)eV = -0.85 eV 

Energy states for Hydrogen
En = - (13.60 / n²) 

n² = -13.60 / -0.85 = 16
n = 4