A hydrogen like atom (atomic number Z) is in a higher excited sate of quantum number n .This excited atom can make a transition to the first excited state by successively emitting two photon of energies `10.20 eV` and `17.00 eV` .Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photon of energy `4.25 ev` and `5.95 eV` Determine the followings:
The value of atomic number (Z) is

Options

A. 2
B. 3
C. 4
D. 5

Answer

B

Explanation

For the first excited state n=2
Total energy emitted for the transition form n = 6 to n =2 is - 13.6 `(1/6^2-1/2^2)Z^2`
and this must be equal to 10.2 + 17.00 = 27.2 eV
`therefore -13.6 (1/36-1/4) Z^2 = 27.2`
`therefore (1/36-1/4) Z^2 = - (27.2)/(13.6) = 2`
`therefore ((4-36)/(144)) Z^2 = -2`
`therefore - (2Z^2)/(9) = -2`
`therefore Z^2 = 9 " or " Z=3`