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 second excited state by emitting two photons of energies 4.24 eV and 5.95 eV. The hydrogen-like atom is :

• A. Li${}^{2+}$
• B. He${}^{+}$
• C. H
• D. none

Answer 1

Answer: (A)

Li${}^{2+}$

$\Delta E=13.6\times Z^2[\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}]$

The electron makes a transition from state $n_2$ is the first excited state $(n_1=2)$ releasing $10.2+17=27.2eV$ energy.

$27.2=Z^2\times 13.6[\frac{1}{4}-\frac{1}{{n_2}^2}]$........(i)

When the electron comes to the second excited state $(n_1=3)$ the energy is $4.25+5.95=10.2eV$

$10.2=Z^2\times 13.6[\frac{1}{9}-\frac{1}{{n_2}^2}]$......(ii)

On solving both the equation (i) and (ii), we get $Z=3$ and $n_2=6$

It is $L{i}^{2+}$ ion.

Hence, option A is correct.