Question

Use the Rydberg Equation to calculate the energy in Joules of the transition between n = 7 and n = 3 for the hydrogen atom. Find the frequency in Hz of this transition if the wavelength is 1000nm.

Answer 1

Answer:

The energy of each transition is approximately $1.98\times 10^{-19}\; \rm J$.

The frequency of photons released in such transitions is approximately $3.00\times 10^{14}\; \rm Hz$.

Explanation:

The Rydberg Equation gives the wavelength (in vacuum) of photons released when the electron of a hydrogen atom transitions from one main energy level to a lower one.

• Let $\lambda_\text{vac}$ denote the wavelength of the photon released when measured in vacuum.
• Let $R_\text{H}$ denote the Rydberg constant for hydrogen. $R_\text{H} \approx 1.09678 \times 10^{7}\; \rm m^{-1}$.
• Let $n_1$ and $n_2$ denote the principal quantum number of the initial and final main energy level of that electron. (Both $n_1\!$ and $n_2\!$ should be positive integers; $n_1 > n_2$.)

The Rydberg Equation gives the following relation:

$\displaystyle \frac{1}{\lambda_\text{vac}} = R_\text{H} \cdot \left(\frac{1}{{n_2}^2}} -\frac{1}{{n_1}^2}\right)$.

Rearrange to obtain and expression for $\lambda_\text{vac}$:

$\displaystyle \lambda_\text{vac} = \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)}$.

In this question, $n_1 = 7$ while $n_2 = 3$. Therefore:

$\begin{aligned} \lambda_\text{vac} &= \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)} \\ &\approx \frac{1}{\displaystyle 1.09678 \times 10^{7}\; \rm m^{-1} \cdot \left(\frac{1}{3^2} - \frac{1}{7^2}\right)} \approx 1.0 \times 10^{-6}\; \rm m \end{aligned}$.

Note, that $1.0\times 10^{-6}\; \rm m$ is equivalent to $1000\; \rm nm$. That is: $1.0\times 10^{-6}\; \rm m = 1000\; \rm nm$.

Look up the speed of light in vacuum: $c \approx 3.00\times 10^{8}\; \rm m \cdot s^{-1}$. Calculate the frequency of this photon:

$\begin{aligned} f &= \frac{c}{\lambda_\text{vac}} \\ &\approx \frac{3.00\times 10^{8}\; \rm m\cdot s^{-1}}{1.0\times 10^{-6}\; \rm m} \approx 3.00 \times 10^{14}\; \rm Hz\end{aligned}$.

Let $h$ represent Planck constant. The energy of a photon of wavelength $f$ would be $E = h \cdot f$.

Look up the Planck constant: $h \approx 6.62607 \times 10^{-34}\; \rm J \cdot s$. With a frequency of $3.00\times 10^{14}\; \rm Hz$ ($1\; \rm Hz = 1\; \rm s^{-1}$,) the energy of each photon released in this transition would be:

$\begin{aligned}E &= h \cdot f \\ &\approx 6.62607 \times 10^{-34}\; \rm J\cdot s^{-1} \times 3.00 \times 10^{14}\; \rm s^{-1} \\ &\approx 1.98 \times 10^{-19}\; \rm J\end{aligned}$.