Question

The wave number of the last line of the Balmer series in the hydrogen spectrum will be ($R = 10^7\text{ m}^{-1}$)

• A. $16 \times 10^4\text{ m}^{-1}$
• B. $8 \times 10^5\text{ m}^{-1}$
• C. $0.25 \times 10^7\text{ m}^{-1}$
• D. $25 \times 10^5\text{ m}^{-1}$

Hint:

The "last line" of any spectral series always has the shortest wavelength and maximum energy because it originates from $n_2 = \infty$. This simplifies the bracketed term in the Rydberg formula to just $\frac{1}{n_1^2}$. For the Balmer series, it is always simply $\frac{R}{4}$.

Answer 1

The Correct Option is C

Step 1: What is asked.
The Balmer series is a set of spectral lines of the hydrogen atom. We need the wave number of its last line, also called the series limit. Wave number means how many waves fit in one metre.
Step 2: The Rydberg formula.
For hydrogen, the wave number $\bar{\nu}$ of a line is \[ \bar{\nu} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \] For the Balmer series the lower level is always $n_1 = 2$.
Step 3: Meaning of the last line.
The lines crowd closer and closer as the electron starts from higher and higher orbits. The very last line comes from the highest orbit of all, which is $n_2 = \infty$.
Step 4: Put the values in.
\[ \bar{\nu} = R\left(\frac{1}{2^2} - \frac{1}{\infty^2}\right) \]
Step 5: Simplify.
Since $1/\infty^2 = 0$, this becomes \[ \bar{\nu} = R \times \frac{1}{4} = \frac{R}{4} \]
Step 6: Use the given R.
With $R = 10^7$ m$^{-1}$: \[ \bar{\nu} = \frac{10^7}{4} = 0.25 \times 10^7\ \text{m}^{-1} \] This is option (3). \[ \boxed{\bar{\nu} = 0.25 \times 10^7\ \text{m}^{-1}} \]