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 excited state by successively emitting two photons of energies $10.2eV$ and $17.0eV$, respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies $4.25eV$ and $5.95eV$, respectively. The values of $n$ and $Z$ are, respectively
A. 6 and 6
B. 3 and 3
C. 6 and 3
D. 3 and 6

Answer 1

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

Formula Used:
As we know the formula for the energy associated with an electron-$13.6{Z^2} \times \left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$.
By substituting the value of ${n_1}$ and ${n_2}$ we can make two different equations and by solving them we can easily determine the value of n and Z.

Complete step by step answer:
As we know that energy associated of an electron, $13.6{Z^2} \times \left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
$E$= $13.6{Z^2} \times \left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{n^2}}}} \right)$
If the transition takes place from excited state $n$ to first excited state${n_1}$=$2$
$\left( {10.20 + 17.00} \right)eV$= $13.6{Z^2} \times \left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{n^2}}}} \right)$ …….. (1)
If the transition takes place from excited state $n$ to second excited state ${n_1} = 3$
$\left( {4.24 + 5.95} \right)eV = 13.6{Z^2} \times \left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{n^2}}}} \right)$ ……… (2)

Now, Dividing the equation (1) by equation (2), we get-
$1.18 = \dfrac{{{n^2} - 4}}{{{n^2} - 9}}$
$0.18{n^2} = 6.6$
${n^2} = 36$
Now take squaring on both sides of the equation, we get a value of n is 6.
$n = 6$
When we Put value of $n = 6$ in equation (1), we will get the value of z,
$Z = 3$
Thus, the value of $n$ and $Z$ are $6$ and $3$.

Hence, option C is the correct answer.

Note: 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.First excited state ${n_1}$=2 and second excited state ${n_2}$ is 3.