Question

The Lyman series corresponds to the emission lines of hydrogen atom due to the transitions of the type $n \rightarrow 1$ ($2 \rightarrow 1, 3 \rightarrow 1, 4 \rightarrow 1$, etc) where $n$ is the principal quantum number of the excited state. The ionization energy of the hydrogen atom can be determined by plotting:

• A. Frequency differences between successive emission lines vs emission frequencies
• B. Emission frequency vs $n^2$
• C. Emission frequency vs $1/n^2$
• D. Frequency differences between successive emission lines vs $n$

Answer 1

Answer: (C)

Emission frequency vs $1/n^2$

Answer 2

The energy of emitted photons in the Lyman series ($n \rightarrow 1$) is $h\nu = 13.6 \left(1 - \frac{1}{n^2}\right)$ eV. Rearranging gives $\nu = \frac{13.6}{h} - \frac{13.6}{h} \frac{1}{n^2}$. This is a linear equation of the form $y = c - mx$, where $y = \nu$ and $x = \frac{1}{n^2}$. The y-intercept ($c$) is $\frac{13.6}{h}$. When $1/n^2 = 0$ (corresponding to $n=\infty$), the frequency is $\nu_{max} = \frac{13.6}{h}$. The energy associated with this frequency, $h\nu_{max} = 13.6$ eV, is the ionization energy of the hydrogen atom. Thus, plotting emission frequency vs $1/n^2$ yields a straight line whose y-intercept gives the frequency corresponding to the ionization energy.