the first lyman transition in the hydrogen spectrum has /\E = 10.2 The same energy change is observed in the second balmer transition ??
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For He+ ion, we have
1/λ = Z2RH [1/n12 -1/n22]
= (2)2RH [1/(2)2 – 1/(4)2] = RH 3/4 ….(i)
Now for hydrogen atom 1/λ = RH [1/n12 -1/n22] ….(ii)
Equating equation (i) and (ii), 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.
1/λ = Z2RH [1/n12 -1/n22]
= (2)2RH [1/(2)2 – 1/(4)2] = RH 3/4 ….(i)
Now for hydrogen atom 1/λ = RH [1/n12 -1/n22] ….(ii)
Equating equation (i) and (ii), 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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