For He+ , a transition takes place from the orbit of radius 105.8 pm to the orbit of radius 26.45 pm.The wavelength (in A0) of the emitted photon during the transition is x×102 then x [Use : Bohr radius, a = 52.9 pm; Rydberg constant , RH=2.2×10−18J ; Planck’s constant , h=6.6×10−34Js ; Speed of light , c=3×108ms−1 ]

For single electron system, rn=52.9×n2Zpm 105.8=52.9×n122∴n12=4⇒n1=226.45=52.9×n222∴n2=1So, transition is from 2 to 1.Now hcλ=RHZ2(1n12−1n22)λ=30×10−9m=30nm

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For $H e^{+}$ , a transition takes place from the orbit of radius 105.8 pm to the orbit of radius 26.45 pm.
The wavelength (in $\overset{0}{A}$ ) of the emitted photon during the transition is $x \times 10^{2}$ then $x$
[Use : Bohr radius, a = 52.9 pm; Rydberg constant , $R_{H} = 2.2 \times 10^{- 18} J$ ; Planck’s constant , $h = 6.6 \times 10^{- 34} J s$ ; Speed of light , $c = 3 \times 10^{8} m s^{- 1}$ ]

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Detailed Solution

For single electron system, $r_{n} = 52.9 \times \frac{n^{2}}{Z} p m$
$105.8 = \frac{52.9 \times n_{1}^{2}}{2} ∴ n_{1}^{2} = 4 \Rightarrow n_{1} = 2$
$26.45 = \frac{52.9 \times n_{2}^{2}}{2} ∴ n_{2} = 1$
So, transition is from 2 to 1.
$\begin{array}{l} N o w \frac{h c}{λ} = R_{H} Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) \\ λ = 30 \times 10^{- 9} m = 30 n m \end{array}$