First slide

Hydrogen spectrum

Question

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 $L i^{+ +}$ will be (Ionization energy of hydrogen atom is 13.6 eV)

Moderate

A

$182.51 Å$
B

$177.17 Å$
C

$142.25 Å$
D

$113.74 Å$

Solution

$E_{n} = - 13.6 \frac{Z^{2}}{n^{2}} eV$
Required energy for said transition,
$Δ E = E_{3} - E_{1} = 13.6 Z^{2} [\frac{1}{1^{2}} - \frac{1}{3^{2}}] \Rightarrow Δ E = 13.6 \times 3^{2} [\frac{8}{9}] = 108.8 eV ΔE = 108.8 \times 1.6 \times 10^{- 19} J Now ΔE = \frac{hc}{λ} = 108.8 \times 1.6 \times 10^{- 19} \Rightarrow λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{108.8 \times 1.6 \times 10^{- 19}} = 0.11374 \times 10^{- 7} m = 113.74 Å$

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