Question

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->2`
`2)` `n = 5->2`
`3)` `n = 7->2`
`4)` `n = 2->1`

Answer 1

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

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

Looking at `n = 2->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->2`. From the Rydberg equation:

`DeltaE = -hcR_H(1/n_f^2 - 1/n_i^2)`

`DeltaE_(4->2) = -(6.626 xx 10^(-34) "J"cdot"s")(2.998 xx 10^(8) "m/s")("109737 m"^(-1))(1/2^2 - 1/4^2)`

`= 4.087 xx 10^(-21) "J"`

`DeltaE_(2->1) = -(6.626 xx 10^(-34) "J"cdot"s")(2.998 xx 10^(8) "m/s")("109737 m"^(-1))(1/1^2 - 1/2^2)`

`= 1.635 xx 10^(-20) "J"`

In fact, it is NOT. So actually, `DeltaE_(2->1) > DeltaE_(4->2)`. Checking the rest:

`DeltaE_(5->2) = -(6.626 xx 10^(-34) "J"cdot"s")(2.998 xx 10^(8) "m/s")("109737 m"^(-1))(1/2^2 - 1/5^2)`

`= 4.578 xx 10^(-21) "J"`

`DeltaE_(7->2) = -(6.626 xx 10^(-34) "J"cdot"s")(2.998 xx 10^(8) "m/s")("109737 m"^(-1))(1/2^2 - 1/7^2)`

`= 5.005 xx 10^(-21) "J"`

Thus, the energy difference ordering is:

`bb(DeltaE_(2->1) > DeltaE_(7->2) > DeltaE_(5->2) > DeltaE_(4->2))`

or:

`(3) > (4) > (2) > (1)`

Thus, the answer is option (3). Always check your numbers.