2142)
Only the energy corresponding to transition for n = 4 to n = 3 falls in visible region.
So λ=hc2.64eV=4.8 *10-7
Hence option C
3 Answers
21
3041) Excitation energy of H atom = -3.4eV + 13.6eV [n2 - n1]
= 9.2eV
Energy of He* in excited state = -13.6 * z2n2
= -13.6* 4/4 = -13.6eV
Energy of He* after transfer of excitation energy = -13.6eV + 9.2eV
= -3.4eV
-3.4eV = -13.6 * 4n2
n2 = 54.43.4 = 16
n = 4