Question

Of the following transitions in the Bohr hydrogen atom, the _______  transition results in the emission of the lowest-energy photon.

When the electron in a hydrogen atom moves from n = 6 ton = 2, light with a wavelength of nm is emitted.

Answer 1

Lowest energy transition will be for n= 2 to n= 1 which happens for emission. And n= 1 ti n= 2 which happens for absorption. This govern by the equation En = (-13.6/n^2) e.v.

By calculating energy gap of two state we can calculate required energy for transition.

2) for n= 2 state E2 = (-13.6/2^2) e.v. for n=6 state E6 = (-13.6/6^2) e.v. ; so energy gap ∆E= E6-E2=3.8 e.v.= 3.02*1.6*10^(-19) J ; this ∆ E should be equal to (hc/wavelength); where h = Planck constant, c = light speed;

Hence wavelength = hc/ ∆E =(6.626*10^-34 * 3*10^8 /3.02*1.6*10^-19) m = 4.11*10^-7 m =411 nm