calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from n=7 to n=1. Is it just 6? or how do I do this?

Answer 1

No.
delta E = 2.180 x 10^-18 J x (1/n1^2 - 1/n2^2)
n1 = 1 and 1^2 = 1
n2 = 7 and 7^2 = 49
Solve for delta E.

Answer 2

2.136 x 10^-18