Question

) Describe the Bohr theory of the hydrogen atom and how it explains the line spectrum of hydrogen. What is the energy and wavelength of the photon of energy that was emitted by hydrogen if one of its electrons dropped from the n=6 state to the n=2 state. In what region of the electromagnetic spectrum does this wavelength fall? Which color line is this in the line spectra of hydrogen?

Answer 1

1)
Here photon will be emitted
1/lambda = R* (1/nf^2 - 1/ni^2)
R is Rydberg constant. R = 1.097*10^7

1/lambda = R* (1/nf^2 - 1/ni^2)
1/lambda = 1.097*10^7* (1/2^2 -   1/6^2)
lambda = 4.102*10^-7 m
lambda = 410 nm
Answer: wavelength = 410 nm

2)
use:
E = h*c/lambda
=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(4.102*10^-7 m)
= 4.846*10^-19 J
Since this is emitted energy, sign will be negative
So,
E = 4.85*10^-19 J
Answer: energy = 4.85*10^-19 J

3)
It falls in visible region

4)
color is blue