Question

For ( mathrm{He}^{+} )
[
begin{aligned}
frac{1}{lambda} &=2^{2} R_{H}left[frac{1}{n_{1}},-frac{1}{n_{2}}right]
&=(2)^{2} R_{H}left[frac{1}{(2)^{2}}-frac{1}{(4)^{2}}right]=R_{H} frac{3}{4} ldots(1)
end{aligned}
]
For ( H ) atom,
[
begin{array}{c}
frac{1}{lambda}=R_{H}left(frac{1}{n_{1}^{2}}-frac{1}{n_{2}^{2}}right] ldots-(11)
(1)=(11), text { we gut }
frac{1}{n_{1}}-frac{1}{n_{2}}=frac{3}{4}
Rightarrow n_{1}=1 text { & } n_{2}=2
end{array}
]
( therefore ) transition ( n=2 ) to ( n=1 ) in ( mathrm{Hatum} )
will bave same warelength ay tranition, ( n=4 ) te ( n=2 ) in ( H e^{t} )

# Example 30. What transition in the hydrogen spectrum have the same wavelength as Balmer transition, n = 4 10 n=2, of Het spectrum? (I.L.T. 1993)

Solution