Answer:
Approximately [tex]2.47\times 10^{15}\; \rm Hz[/tex].
Explanation:
The Lyman Series of a hydrogen atom are due to electron transitions from energy levels [tex]n \ge 2[/tex] to the ground state where [tex]n = 1[/tex]. In this case, the electron responsible for the line started at [tex]n = 2[/tex] and transitioned to
A hydrogen atom contains only one electron. As a result, Bohr Model provides a good estimate of that electron's energy at different levels.
In Bohr's Model, the equation for an electron at energy level [tex]n[/tex] (
[tex]\displaystyle - \frac{k\, Z^2}{n^2}[/tex] (note the negative sign in front of the fraction,)
where
The electron that produced the [tex]n = 2[/tex] line was initially at the
[tex]\begin{aligned} &E_{n = 2} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{2^2} \cr & \approx -5.4475\times 10^{-19}\; \rm J\end{aligned}[/tex].
The electron would then transit to energy level [tex]n = 1[/tex]. Its energy would become:
[tex]\begin{aligned} &E_{n = 1} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{1^2} \cr & \approx -2.179 \times 10^{-18} \; \rm J\end{aligned}[/tex].
The energy change would be equal to
[tex]\begin{aligned}&\text{Initial Energy} - \text{Final Energy} \cr &= E_{n = 2} - E_{n = 1} \cr &= -5.4475 \times 10^{-19} - \left(-2.179 \times 10^{-18}\right) \cr & \approx 1.63425\times 10^{-18}\; \rm J \end{aligned}[/tex].
That would be the energy of a photon in that [tex]n = 2[/tex] spectrum line. Planck constant [tex]h[/tex] relates the frequency of a photon to its energy:
[tex]E = h \cdot f[/tex], where
In this case, [tex]E \approx 1.63425 \times 10^{-18}\; \rm J[/tex]. Hence,
[tex]\begin{aligned} f &= \frac{E}{h} \cr &\approx \frac{1.63425\times 10^{-18}}{6.62607015\times 10^{-34}} \cr & \approx 2.47 \times 10^{15}\; \rm s^{-1}\end{aligned}[/tex].
Note that [tex]1 \; \rm Hz = 1 \; \rm s^{-1}[/tex].
