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

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

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

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Answer 1 · 2017-03-26T04:11:50

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

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

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

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

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

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

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

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

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