Question

In Li⁺⁺, electron in first Bohr orbit is excited to a level by a radiation of wavelength λ. When the ion gets deexcited to the ground state in all possible ways(including intermediate emission) a total of six spectral lines are observed. What is the value of λ?
(Given: h = 6.63 × 10³⁴ js; e = 3 × 10⁸ ms⁻¹)
(A) 10.8 nm (B) 11.4 nm
(C) 9.4 nm (D) 12.3 nm

Answer 1

answer : option (A) 10.8 nm

given, total number of spectral lines are observed = 6

⇒n(n - 1)/2 = 6

⇒n² - n - 12 = 0

⇒n² - 4n + 3n - 12 = 0

⇒n = 4, - 3

so, n = 4

hence transmission from n₁ = 1 to n₂ = 4

now energy , ∆E = 13.6 × Z²[1/n₁² - 1/n₂² ]

= 13.6 × (3)² [1/1² - 1/4²]

= 13.6 × 9 × 15/16

= 114.75 eV

now using formula, ∆E = hc/λ

λ = 1240/∆E (in eV) nm

= 1240/114.75 = 10.8 nm

hence option (A) is correct choice..