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Question: In \(L{i^{ + + }}\), electrons in the first Bohr orbit are excited to a level by radiation of wavele...

In Li++L{i^{ + + }}, electrons in the first Bohr orbit are excited to a level by radiation of wavelength λ\lambda . When the ion gets deexcited to the ground state in all possible ways (including intermediate emissions), a total of 6 spectral lines are observed. What is the value of λ\lambda ?
(Given: h=6.63×1034Jsh = 6.63 \times {10^{34}}Js; c=3×108m/sc = 3 \times {10^8}m/s)
(a) 9.4 nm
(b) 12.3 nm
(c) 10.8 nm
(d) 11.4 nm

Explanation

Solution

The spectral lines observed can be used to find the transition level of the atom. As transition energies are fixed for energy levels, this can be used to determine λ\lambda .

Formula used:
Number of spectral lines is given by:
No. of spectral lines=n(n1)2 = n\dfrac{{(n - 1)}}{2} ……(1)
where n is the transition level.
Change in energy for transition between energy levels n1{n_1} and n2{n_2} is given by:
ΔE=13.6×Z2(1n121n22)\Delta E = 13.6 \times {Z^2}(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}) ……(2)
where Z is the atomic number of the atom.
Energy is related to wavelength by:
ΔE=hc/λ\Delta E = hc/\lambda ……(3)

Complete step-by-step answer:

Given:
1. Atom is Li++L{i^{ + + }} with Z=3Z = 3.
2. No. of spectral lines = 6.

To find: The wavelength λ\lambda which excited the atom.

Step 1 of 3:
Find the energy level transition from no. of spectral lines using eq (1):
6=n(n1)2 n2n12=0 (n4)(n+3)=0 n=4,3 \begin{gathered} 6 = n\dfrac{{(n - 1)}}{2} \\\ {n^2} - n - 12 = 0 \\\ (n - 4)(n + 3) = 0 \\\ n = 4, - 3 \\\ \end{gathered}
As n cannot be negative, n is 4. This means the transition occurred from n1=1{n_1} = 1 to n2=4{n_2} = 4.

Step 2 of 3:
Find ΔE\Delta E for transition from n1=1{n_1} = 1 to n2=4{n_2} = 4 using eq (2):
ΔE=13.6×32(112142) ΔE=13.6×9×1516 ΔE=114.75eV \begin{gathered} \Delta E = 13.6 \times {3^2}(\dfrac{1}{{{1^2}}} - \dfrac{1}{{{4^2}}}) \\\ \Delta E = 13.6 \times 9 \times \dfrac{{15}}{{16}} \\\ \Delta E = 114.75eV \\\ \end{gathered}

Step 3 of 3:
Find λ\lambda by putting hc=1240eV.nmhc = 1240eV.nm in eq (3):
114.75eV=1240eV.nm/λ λ=10.8nm \begin{gathered} 114.75eV = 1240eV.nm/\lambda \\\ \lambda = 10.8nm \\\ \end{gathered}

Correct Answer:
The wavelength λ\lambda which excited the atom: (c) 10.8 nm

Note: In questions like these first find the transition level. Then find the corresponding energy for this transition. From this energy, we can find the wavelength.