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PROBLEM

3
8 points
In units of hRh R, what is the amount of energy associated with the transition from n=4n=4 to n=1n=1 in the hydrogen emission spectrum? ( h is Planck's constant and R=3.3×1015 HzR=3.3 \times 10^{15} \mathrm{~Hz} in the Rydberg equation, although you don't need to use these numbers in this problem.)
116hR\frac{1}{16} h R
34hR\frac{3}{4} h R
14hR\frac{1}{4} h R
1516hR\frac{15}{16} h R
4
8 points

STEP 1

What is this asking?
How much energy is released when an electron in a hydrogen atom jumps from a higher energy level (n=4) to a lower energy level (n=1), expressed in terms of hR?
Watch out!
Don't forget that the energy change is negative since energy is released!
But we're asked for the amount of energy, which is positive.

STEP 2

1. Rydberg Formula
2. Calculate the Energy Change

STEP 3

Alright, let's start with the Rydberg formula, which describes the energy change when an electron transitions between energy levels in a hydrogen atom.
It's like a secret code for the atom's energy!
The formula is:
ΔE=hR(1nf21ni2) \Delta E = -hR \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) Where ΔE\Delta E is the change in energy, h is Planck's constant, R is the Rydberg constant, nin_i is the initial energy level of the electron, and nfn_f is the final energy level.

STEP 4

Remember, a negative ΔE\Delta E means energy is released, while a positive ΔE\Delta E means energy is absorbed.
We're dealing with emission here, so we expect a negative value for ΔE\Delta E.

STEP 5

Let's plug in our values.
We're given ni=4n_i = 4 (the initial level) and nf=1n_f = 1 (the final level).
So, we have:
ΔE=hR(112142) \Delta E = -hR \left( \frac{1}{1^2} - \frac{1}{4^2} \right)

STEP 6

Now, let's simplify those fractions:
ΔE=hR(11116) \Delta E = -hR \left( \frac{1}{1} - \frac{1}{16} \right) ΔE=hR(1116) \Delta E = -hR \left( 1 - \frac{1}{16} \right)

STEP 7

To subtract the fractions, we need a common denominator, which is 16:
ΔE=hR(1616116) \Delta E = -hR \left( \frac{16}{16} - \frac{1}{16} \right)

STEP 8

Now, subtract:
ΔE=hR(16116) \Delta E = -hR \left( \frac{16 - 1}{16} \right) ΔE=hR(1516) \Delta E = -hR \left( \frac{15}{16} \right) ΔE=1516hR \Delta E = -\frac{15}{16} hR

STEP 9

The energy change is 1516hR-\frac{15}{16} hR.
Since the question asks for the amount of energy, we take the absolute value, making it positive:
ΔE=1516hR |\Delta E| = \frac{15}{16} hR

SOLUTION

The amount of energy associated with the transition is 1516hR\frac{15}{16} hR.

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