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 the second transition is 3. The value of atomic number (Z) is _______.
(A) 2
(B) 4
(C) 6
(D) 3

Answer 1

Hint : The number of protons in the nucleus of an atom determines the atomic number of an element. As we know that the hydrogen atom is the simplest in nature. It consists of a single negatively charged electron that moves about a positively charged proton. Here transition is happening from first to a second excited state, so in the energy formula it remembers to substitute the values of first and second excited states respectively.

Complete Step By Step Answer:
As we know the formula for the energy associated with electron:
$E=13.6{{Z}^{2}}\times (\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}})\,\,\,\,.....(1)$
By substituting the value of ${{n}_{1}}$ and ${{n}_{2}}$ we can make two different equations and by solving them we can determine the value of Z.
If the transition takes place from excited state $n$ to first excited state ${{n}_{1}}=2$ ,
$(10.20\,+\,17.00)\,eV=13.6{{Z}^{2}}\times (\dfrac{1}{{{2}^{2}}}-\dfrac{1}{n_{2}^{2}})\,\,\,\,.....(2)$
If the transition takes place from excited state $n$ to first excited state ${{n}_{2}}=3$ ,
$(4.24\,+\,5.95)\,eV=13.6{{Z}^{2}}\times (\dfrac{1}{{{3}^{2}}}-\dfrac{1}{n_{2}^{2}})\,\,\,\,.....(3)$
Now, dividing the equation (2) by equation (3), we get:
$1.18=\dfrac{n_{2}^{2}-4}{n_{2}^{2}-9}$
$n_{2}^{2}=36$
Taking square-root on both sides, we get:
${{n}_{2}}=6$
Now substituting the value of ${{n}_{2}}$ in equation (2), we get:
$(10.20\,+\,17.00)\,eV=13.6{{Z}^{2}}\times (\dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{6}^{2}}})$
On calculating the above equation we get:
$Z=3$
Hence the value of the atomic number is 3. Hence option D. is correct.

Note :
In order to solve this kind of question we should have knowledge about Bohr’s model. In Bohr’s model, the electron is pulled around the proton in a circular orbit by an attractive Coulomb force. The excited state describes an atom, ion, or molecule with an electron in a higher than normal energy level than its ground state.