# An electron in a hydrogen atom drops from energy level n=5 to n=3. What is the energy transition using the Rydberg equation?

Mar 23, 2015

The energy transition will be equal to $1.55 \cdot {10}^{- 19} \text{J}$.

So, you know your energy levels to be n = 5 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition

$\frac{1}{l a m \mathrm{da}} = R \cdot \left(\frac{1}{n} _ {\left(\text{final")^(2) - 1/n_("initial}\right)}^{2}\right)$, where

$l a m \mathrm{da}$ - the wavelength of the emitted photon;
$R$ - Rydberg's constant - $1.0974 \cdot {10}^{7} {\text{m}}^{- 1}$;
${n}_{\text{final}}$ - the final energy level - in your case equal to 3;
${n}_{\text{initial}}$ - the initial energy level - in your case equal to 5.

So, you've got all you need to solve for $l a m \mathrm{da}$, so

$\frac{1}{l a m \mathrm{da}} = 1.0974 \cdot {10}^{7} {\text{m}}^{- 1} \cdot \left(\frac{1}{3} ^ 2 - \frac{1}{5} ^ 2\right)$

$\frac{1}{l a m \mathrm{da}} = 0.07804 \cdot {10}^{7} \text{m"^(-1) => lamda = 1.28 * 10^(-6)"m}$

Since $E = \frac{h c}{l a m \mathrm{da}}$, to calculate for the energy of this transition you'll have to multiply Rydberg's equation by $h \cdot c$, where

$h$ - Planck's constant - $6.626 \cdot {10}^{- 34} \text{J" * "s}$
$c$ - the speed of light - $\text{299,792,458 m/s}$

So, the transition energy for your particular transition (which is part of the Paschen Series) is

E = (6.626 * 10^(-34)"J" * cancel("s") * "299,792,458" cancel("m/s"))/(1.28 * 10^(-6)cancel("m"))

$E = 1.55 \cdot {10}^{- 19} \text{J}$

Mar 24, 2015

A photon of energy $1.531 \times {10}^{- 19} \text{J}$ will be emitted.

The Rydberg expression for an electronic transition in the hydrogen atom is:

$\frac{1}{\lambda} = R \left[\frac{1}{n} _ {1}^{2} - \frac{1}{n} _ {2}^{2}\right]$

$\lambda = \text{wavelength}$

$R = 1.097 \times {10}^{7} {m}^{- 1}$

${n}_{1} = 3$

${n}_{2} = 5$

So:

$\frac{1}{\lambda} = R \left(\frac{1}{9} - \frac{1}{25}\right)$

$\frac{1}{\lambda} = R \left(0.071\right) = 1.097 \times {10}^{7} \times 0.071 = 7.789 \times {10}^{5} {m}^{- 1}$

The energy of the photon given by this transition is given by:

$E = h f$

Since $c = f \lambda$

$E = \frac{h c}{\lambda}$

$h$ is the Planck Constant = $6.63 \times {10}^{- 34} \text{Js}$

$c$ = the speed of light =$3 \times {10}^{8} \text{m/s}$

So:

$E = 6.63 \times {10}^{- 34} \times 3 \times {10}^{8} \times 7.789 \times {10}^{5} \text{J}$

$E = 1.531 \times {10}^{- 19} \text{J}$