Question

Calculate the energy of a photon produced when an electron in hydrogen makes a transition from an excited state in the 5th n=5 level, back to the ground state, n=1. Would you expect to see radiation from this photon? Explain.

Answer 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* (1/1^2 - 1/5^2)

lambda = 9.496*10^-8 m

lambda = 95 nm

use:

E = h*c/lambda

=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(9.496*10^-8 m)

= 2.093*10^-18 J

Since this is emitted energy, sign will be negative

So,

E = 2.09*10^-18 J

Answer: 2.09*10^-18 J

wavelength here is 95 nm which do not fall in visible region

So, we can’t see this radiation