A hydrogen-like atom is in a higher energy level of quantum number 6. The excited atom make a transition to first excited state by emitting photons of total energy 27.2 eV. The atom from the same excited state make a transition to the second excited state by successively emitting two photons. If the energy of one photon is 4.25 eV, find the energy of other photon.

Options

A. 5.95 eV
B. 6.25 eV
C. 6.95 eV
D. 7.80 eV

Answer

A

Explanation

`DeltaE=13.6 Z^2[1/n_1^2-1/n_2^2]`
Given, `n_2=6` and for first excited state `n_1=2`
`:. 27.2 = 13.6 Z^(2) [1/4 -1/36] ------(1)`
`4.25+x=13.6 Z^(2) [1/9 -1/36]` where `n_1 = 3 ----(2)`
By dividing eqn. (1) and eqn. (2)
`(27.2)/(4.25 +x)=4/18 xx12 =8/3 :. x = 5.95 eV`