the frequency of light emitted for the transition n=4 to n=2 of helium

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

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

Answer»

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.

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