Question

In a certain electronic transition from the quantum ‘n’ to the ground state in atomic H in one or more steps, no line belonging to the bracket series is observed. What wave number may be observed in the Balmer series?
(a)- $\dfrac{8R}{9},\dfrac{5R}{36}$
(b)- $\dfrac{3R}{16},\dfrac{8R}{9}$
(c)- $\dfrac{5R}{36},\dfrac{3R}{16}$
(d)- $\dfrac{3R}{4},\dfrac{3R}{16}$

Answer 1

The spectrum of hydrogen contains the Lyman series, Balmer series, Paschen series, Brackett series, and Pfund series, so the lowest quantum number for Lyman series is 1, Balmer series is 2, Paschen series is 3, Brackett series is 4, and Pfund series is 5. We can use the formula:
$\bar{v}=R\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$

Complete answer: The spectrum of hydrogen contains the Lyman series, Balmer series, Paschen series, Brackett series, and Pfund series, so the lowest quantum number for Lyman series is 1, Balmer series is 2, Paschen series is 3, Brackett series is 4, and Pfund series is 5.
So, we can see that the value of ‘n’ must be 4, if it is greater than 4, then the Brackett series will be noticed.
Let us calculate the total number of lines, by:
$\dfrac{n(n-1)}{2}=\dfrac{4\text{ x 3}}{2}=6$
Therefore, there are 6 lines between these transitions, these are written below:
n = 4 $\to$ n = 3
n = 4 $\to$ n = 2
n = 4 $\to$ n = 1
n = 3 $\to$ n = 2
n = 3 $\to$ n = 1
n = 2 $\to$ n = 1
So, we have to find the wavenumber for the Blamer series, and for the Balmer series the value of n is 2. Therefore, from the above transitions, only two transitions will belong to Balmer series, these are:
n = 4 $\to$ n = 2
n = 3 $\to$ n = 2
Now, we can calculate the wavenumber by:
$\bar{v}=R\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$
${{\bar{v}}_{1}}=R\left[ \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{4}^{2}}} \right]=\dfrac{3}{16}R$
${{\bar{v}}_{2}}=R\left[ \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right]=\dfrac{5}{36}R$
Therefore, the wave numbers are $\dfrac{5R}{36},\dfrac{3R}{16}$.

Hence, the correct answer is an option (c).

Note: Whenever you are using the formula $\bar{v}=R\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$, the value of ${{n}_{1}}$ will be smaller and the value of ${{n}_{2}}$ should be larger, if you exchange them, then the answer will be incorrect.