The n2 value of transition in the He+ spectrum which has the same wavelength as the first Lyman transition of hydrogen (n = 2 to n = 1) ?

At. no. (Z) of He= 2. RH= 109678 cm- . Given n1 = 1, n2 = 2. We know that:              v¯H=1λH=1096781n12−1n22Z2=109678112−122×12cm−=109678×34cm−∴ λH=4109678×3cm.As   ΔEHe+=Z2EH′ similarly, λHe+=ZHe2×λHor     λHe+=(2)2×4109678×3cm;        1λHe+=109678×316cm−For He+, let n1=2, n2=?. Thus we have :             μ¯He+=1λHe+=109678122−1n22cm−or 109678×316cm−=10967814−1n22cm−;                           316=14−1n221n22=14−316=116 ; n22=16.Hence               n2=(16)1/2=4.