Question

Calculate the wavelength, λ, in meters of a photon capable of exciting an electron in He+ from the ground state to n = 4.

Answer 1

\[\lambda =\frac{ 912 (n1^2n^2) }{ z^2 * (n2^2-n1^2 }\]

Answer 2

n1 = 1

n2 = 4

z= 2

Answer 3

use the rydberg equation:

\[\frac{ 1 }{ \lambda }=RZ ^{2}(\frac{ 1 }{ n _{f}^{2} }- \frac{ 1 }{ n _{i}^{2} } )\]

lambda is the wavelength
R = rydgberg constant = 1.09678 x 10^7 /m
Z= number of protons
nf= final level of principal quantum number (in your case 4)
ni=initial level (in your case ground state)

notice that your answer will be in meters and you'll have to convert to nm (which is the unit is which wavelength is expressed as in the visible spectrum (which is the region that excites electrons))