The helium gas in an excited state makes transition from excited state of principal quantum number n = 5 to ground state. The most energetic photons have energy 52.224 eV, find the energy of least energetic photons.

Options

A. 1.224 eV
B. 2.42 eV
C. 3.22 eV
D. 3.82 eV

Answer

A

Explanation

Maximum energy is liberated when transition is from n = 5 to n = 1 and minimum energy is liberated when transition is from n = 5 to n = 4.
`E_1/5^(2)-E_1 = 52.224=E_1 =(-)54.4eV and 5 E_(1)/5^2-E_(1)/4^2 =9/400E_(1) = 9/400xx54.4 eV = 1.22`