The correct option is C 972.5∘A
We know, number of spectral lines = n(n−1)2
where, n is the orbit number.
Six spectral lines are emitted,
n(n−1)2 = 6
n2−n−12=0.
On solving, n = 4, -3 but n = -3 is not possible
Energy in an orbit En = −13.6Z2n2
where, Z = Atomic number
Difference in Energy from ground state to excited state = En=4−En=0
ΔE=−13.6142−(−13.6112) =
ΔE = 13.6 - 0.85 eV = 12.75 eV
According to Planck's quantum theory E = 12400λ eV
where, λ = wavelength of radiation in ∘A
12.75 = 12400λ
λ = 1240012.75 ∘A = 972.5 ∘A