Question

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 state by successively emitting two photons of energies 10.20 eV and 17.00 eV. Alternatively, the atom from the same excited state can make a transition to the energy 4.24 eV and 5.95 eV. If the excited level for second transition is 3.The value of atomic number(Z) is_______.

• A. 2
• B. 4
• C. 6
• D. 3

Answer 1

The correct option is D 3

As we know,

$\Delta E$ = 13.6 $Z^2\times \left(\frac{1}{2^2}-\frac{1}{n^2}\right)$

For the transition from excited state n to first excited state i.e. $n_1$ = 2,
(10.20 + 17.00) eV = 13.6 $Z^2\times \left(\frac{1}{2^2}-\frac{1}{n^2}\right)$ ------ equation (1)
For the transition from excited state n to second excited state i.e. $n_1$ = 3,
(4.24 + 5.95) eV = 13.6 $Z^2\times \left(\frac{1}{3^2}-\frac{1}{n^2}\right)$ ------ equation (2)

On dividing equation 1 and 2,

1.18 = $\frac{n^2-4}{n^2-9}$

0.18 $n^2$ = 6.6

$n^2$ = 36

n = 6

Putting n=6 in equation 1 we get Z=3

Correct answer is D.