Question #17913

Calculate the wavelength, in metres, of a photon capable of exciting an electron in He+ from the ground state to n=4... Someone please tell me the right answer urgently :( ?

Expert's answer

Energy levels for the hydrogen-like He^{+} are given by

E(n) = -Z^{2}*R/n^{2}, where Z = 2;

Energy of an absorbed photon between the levels n=1 and m=4

E(m,n) = Z^{2}*R*(1/n^{2} - 1/m^{2})

We will take R = 10973731.6 m^{-1}, Rydberg constant for H, in reciprocal meters.

Photon wavelength

lambda = h*c/E = 1/Z^{2}*R*(1/n^{2} - 1/m^{2})) = 1/(2^{2}*10,973,731.6 m^{-1}*(1/1^{2} - 1/4^{2})) = 3.746*10^{-8} m

Answer: 3.746*10^{-8} meters

E(n) = -Z

Energy of an absorbed photon between the levels n=1 and m=4

E(m,n) = Z

We will take R = 10973731.6 m

Photon wavelength

lambda = h*c/E = 1/Z

Answer: 3.746*10

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