# How do you determine which of these hydrogen atom energy transitions has the highest energy difference? Isn't it (4) since it is the odd one out?

## 1) $n = 4 \to 2$ 2) $n = 5 \to 2$ 3) $n = 7 \to 2$ 4) $n = 2 \to 1$

Mar 26, 2017

It is actually $\left(3\right)$ that is highest in energy. This is tricky because $\left(4\right)$ does not end on $n = 2$.

Well, you can compare options (1), (2), and (4) and see that since $n$ represents the energy level, an $n = 7 \to 2$ transition is highest in energy compared to $n = 4 \to 2$ and $n = 5 \to 2$.

Looking at $n = 2 \to 1$, we would have to actually check since it isn't towards $n = 2$, but initially one would guess that it is probably less than $n = 4 \to 2$. From the Rydberg equation:

$\Delta E = - h c {R}_{H} \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)$

$\Delta {E}_{4 \to 2} = - \left(6.626 \times {10}^{- 34} {\text{J"cdot"s")(2.998 xx 10^(8) "m/s")("109737 m}}^{- 1}\right) \left(\frac{1}{2} ^ 2 - \frac{1}{4} ^ 2\right)$

$= 4.087 \times {10}^{- 21} \text{J}$

$\Delta {E}_{2 \to 1} = - \left(6.626 \times {10}^{- 34} {\text{J"cdot"s")(2.998 xx 10^(8) "m/s")("109737 m}}^{- 1}\right) \left(\frac{1}{1} ^ 2 - \frac{1}{2} ^ 2\right)$

$= 1.635 \times {10}^{- 20} \text{J}$

In fact, it is NOT. So actually, $\Delta {E}_{2 \to 1} > \Delta {E}_{4 \to 2}$. Checking the rest:

$\Delta {E}_{5 \to 2} = - \left(6.626 \times {10}^{- 34} {\text{J"cdot"s")(2.998 xx 10^(8) "m/s")("109737 m}}^{- 1}\right) \left(\frac{1}{2} ^ 2 - \frac{1}{5} ^ 2\right)$

$= 4.578 \times {10}^{- 21} \text{J}$

$\Delta {E}_{7 \to 2} = - \left(6.626 \times {10}^{- 34} {\text{J"cdot"s")(2.998 xx 10^(8) "m/s")("109737 m}}^{- 1}\right) \left(\frac{1}{2} ^ 2 - \frac{1}{7} ^ 2\right)$

$= 5.005 \times {10}^{- 21} \text{J}$

Thus, the energy difference ordering is:

$\boldsymbol{\Delta {E}_{2 \to 1} > \Delta {E}_{7 \to 2} > \Delta {E}_{5 \to 2} > \Delta {E}_{4 \to 2}}$

or:

$\left(3\right) > \left(4\right) > \left(2\right) > \left(1\right)$