Question

The difference in the energies between the first Balmer lines emitted by A and B is
(A) 12.1 eV
(B) 13.6 eV
(C) 14.3 eV
(D) 15.1 eV

Answer 1

We know that the Balmer line transition occurs from electrons transition from higher to the energy down to the level with principal quantum number $n = 2$.

Formula used: We will calculate the energy difference between the Balmer lines emitted by A and B with the help of the formula $\Delta E = 13.6\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)eV$
Where $\Delta E =$, change in the energy
${n_1}. {n_2}$ are the integers corresponding principal quantum numbers involved in the transition
So, we will calculate the energies of the given transitions and arrive at the required solution.

Complete Answer:
We know that energy difference between the ${n_1}$ and ${n_2}$ orbit,
$\Delta E = 13.6\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)eV$
We assume the atomic numbers of A and B be ${Z_P}$ and ${Z_Q}$ respectively.
So, the first line of transition from ${n_2} = 2$ to ${n_1} = 1$ of atoms A and B is $\Delta {E_\iota } \Rightarrow 81.6 = 13.6\left[ {\dfrac{1}{{{l^2}}} - \dfrac{1}{{{2^2}}}} \right]\left( {Z_P^2 - Z_Q^2} \right)$
$\Rightarrow 13.6(Z_A^2 - Z_B^2) = 108.8eV \to 1$
Now for first Balmer series transition from ${n_2} = 3$ to ${n_1} = 2$ of atoms A and B, we have
$\Delta {E_B} = 13.6\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right) = \dfrac{1}{{36}} \times 108.8$
$\Delta {E_B} = 15.1eV$
Therefore, the difference in energies is 15.11 eV. Hence, the correct solution is option D.

Note: The Balmer transition occurs for the electron transitioning from a higher level to down the energy level with $n = 2$ which is also characterized as ${H_\alpha }$ transition. In the above question, a Lyman transition is also happening for the atoms A and B due to which energy difference is created so keep in mind the different transitions of electrons in the question to calculate the correct solution. The change in energy corresponding to the first line transition of the Lyman series is not given so it should be calculated/memorized to find the difference in energies pertaining to Balmer transitions of atoms A and B.