A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.20 eV and 17.00 eV respectively. Alternatively the atom from the same excited state can make a transition to the second excited state by snccessively emitting two photons of energies 4.25 eV and 5.95 eV respectively. Determine the values of n and Z (ionization energy of hydrogen atom = 13.6 eV)

Options

A. 1
B. 2
C. 3
D. 4

Answer

C

Explanation

Total energy emitted by photo-electron
=`10.2 +17 =27.20 eV`
Since, `E_1 `= Photon of energy emitted through the transition
n= n ton `=2 implies (hc)/(lambda_1) = 27.20 eV`
We have `1/(lambda_1) = R_H . Z^2 (1/(z^2) - 1/(n^2)) ` or
`(hc)/(lambda_1) (hc) R_H . Z^2 (1/(2^2) - 1/(n^2))`
`therefore 27.20 = (hc) R_H . Z^2 (1/4 - 1/n^2) to (1)`
Similarly , total energy liberated during transition of electron from n= n to n = 3 is
`E_2 (hc)/(lambda_2) (4.25 + 5.95) = 10 . 20 eV`
`therefore 10.20 = (hc) R_H Z^2 (1/9 - 1/(n^2)) to (2)`
Dividing Eq. (1) by (2) , we get n = 6 and putting n= 6 in Eq. (1) or (2) , we get , Z = 3