A double charged lithium atom is equivalent to hydrogen whose atomic number is 3. The wavelength of required radiation for emitting electron from first to third Bohr orbit in ${Li}^{+ +}$ will be (Ionisation energy of hydrogen atom is 13.6 eV)

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182.51 $\overset{0}{A}$

177.17 $\overset{0}{A}$

142.25 $\overset{0}{A}$

113.74 $\overset{0}{A}$

answer is D.

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Detailed Solution

$\begin{array}{l} E_{n} = 13 .6 \frac{Z^{2}}{n^{2}} eV . \\ ΔE = E_{3} - E_{1} = 13 .6 Z^{2} [\frac{1}{1^{2}} - \frac{1}{3^{2}}] \\ \Rightarrow ΔE = 13 .6 \times 3^{2} [\frac{8}{9}] = 108 .8 eV \\ \Rightarrow ΔE = 108 .8 \times 1 .6 \times 10^{- 19} J \\ Now ΔE = \frac{hc}{λ} = 108 .8 \times 1 .6 \times 10^{- 19} \\ \Rightarrow λ = \frac{66 \times 10^{- 34} \times 3 \times 10^{8}}{108 .8 \times 1 .6 \times 10^{- 19}} = 0 .11374 \times 10^{- 7} m = 113 .74 \overset{0}{A} \end{array}$