# Question #827e8

Apr 20, 2017

$4.09 \cdot {10}^{- 19} J$

#### Explanation:

Use the formula:
${E}_{n} = - {R}_{H} \left(\frac{1}{n} ^ 2\right)$
where :
${R}_{H} = \text{Rydberg constant} = 2.18 \cdot {10}^{- 18} J$,
$n = \text{the energy level}$

For $n = 4$:
${E}_{4} = - 2.18 \cdot {10}^{- 18} J \left(\frac{1}{4} ^ 2\right) = - 1.3625 \cdot {10}^{- 19} J$

For $n = 2$
${E}_{2} = - 2.18 \cdot {10}^{- 18} J \left(\frac{1}{2} ^ 2\right) = - 5.45 \cdot {10}^{- 19} J$

Then to find the energy of the photon, you just calculate the change in energy.

$\Delta E = {E}_{\text{initial" - E_"final}}$

$\Delta E = \left(- 1.3625 \cdot {10}^{- 19} J\right) - \left(- 5.45 \cdot {10}^{- 19} J\right)$

$\Delta E = 4.0875 \cdot {10}^{- 19} J$

Since you only have 3 significant figures, the answer would be:
$4.09 \cdot {10}^{- 19} J$