# According to the Bohr model for the hydrogen atom, how much energy is necessary to excite an electron from n=1 to n=2?

May 13, 2016

This energy can be determined from the Rydberg equation, which is

$\setminus m a t h b f \left(\frac{1}{\lambda} = {R}_{H} \left(\frac{1}{{n}_{i}^{2}} - \frac{1}{{n}_{j}^{2}}\right)\right)$

where:

• $\lambda$ is the wavelength in units of $\text{m}$.
• ${R}_{H}$ is the Rydberg constant, ${\text{10973731.6 m}}^{-} 1$.
• ${n}_{i}$ is the principal quantum number $n$ for the lower energy level.
• ${n}_{j}$ is the principal quantum number $n$ for the higher energy level.

In this case, ${n}_{i} = 1$ and ${n}_{j} = 2$, since the excitation from $n = 1$ to $n = 2$ is upwards. Thus:

$\frac{1}{\lambda} = {\text{10973731.6 m}}^{-} 1 \left(\frac{1}{1} ^ 2 - \frac{1}{2} ^ 2\right)$

$= {\text{10973731.6 m}}^{-} 1 \left(\frac{1}{1} ^ 2 - \frac{1}{2} ^ 2\right)$

$= {\text{8230298.7 m}}^{- 1}$

Now, the wavelength is

$\textcolor{g r e e n}{\lambda} = \frac{1}{{\text{8230298.7 m}}^{- 1}}$

$= \textcolor{g r e e n}{1.215 \times {10}^{- 7} \text{m}} ,$

or $\text{121.5 nm}$. Let's relate this back to the energy. Recall the equation:

$\setminus m a t h b f \left(\Delta E = h \nu = \frac{h c}{\lambda}\right)$

where:

• $\Delta E$ is the change in energy in $\text{J}$.
• $h$ is Planck's constant, $6.626 \times {10}^{- 34} \text{J"cdot"s}$.
• $\nu$ is the frequency in ${\text{s}}^{- 1}$. Remember that $c = \lambda \nu$.
• $c$ is the speed of light, $2.998 \times {10}^{8} \text{m/s}$.
• $\lambda$ is the wavelength in $\text{m}$ like before.

So, the absorption of energy into a single hydrogen atomic system associated with this process is:

color(blue)(DeltaE) = ((6.626xx10^(-34) "J"cdotcancel"s")(2.998xx10^(8) cancel"m/s"))/(1.215xx10^(-7) cancel"m")

$= \textcolor{b l u e}{1.635 \times {10}^{- 18} \text{J}}$

(absorption is an increase in energy for the system, thus $\Delta E > 0$.)