Question

the frequency of light emitted for the transition n=4 to n=2 of helium + is equal to that of hydrogen atom

Answer 1

The frequency of light emitted for the transition n=4 to n=2 of helium + is equal to that of hydrogen atom will be n = 2 to n = 1.

For He+ ion, we have

1/λ = Z2R¬H [1/n12 -1/n22]

= (2)2RH [1/(2)2 – 1/(4)2] = RH 3/4 (1)

For hydrogen atom 1/lemda = RH [1/n12 -1/n22] (2)

Equating equation (1) and (2), we get

1/n12 -1/n22 = 3/4

Obviously, n1 = 1 and n2 = 2

Hence, the transition n = 2 to n = 1 in hydrogen atom will have the same wavelength as the transition, n = 4 to n = 2 in He+ species.

Answer 2

Answer:

n=2 n=1

Explanation:

hope it helps...