Math / Algebra

Question3 8 points
In units of $h R$ , what is the amount of energy associated with the transition from $n=4$ to $n=1$ in the hydrogen emission spectrum? ( h is Planck's constant and $R=3.3 \times 10^{15} \mathrm{~Hz}$ in the Rydberg equation, although you don't need to use these numbers in this problem.) $\frac{1}{16} h R$ $\frac{3}{4} h R$ $\frac{1}{4} h R$ $\frac{15}{16} h R$ 4 8 points

Studdy Solution

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: $\Delta E = -hR \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$ Where $\Delta E$ is the change in energy, h is Planck's constant, R is the Rydberg constant, $n_i$ is the initial energy level of the electron, and $n_f$ is the final energy level.

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

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Studdy Solution

STEP 1

STEP 2

1. Rydberg Formula
2. Calculate the Energy Change

STEP 3

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

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Studdy Solution

STEP 1

STEP 2

1. Rydberg Formula2. Calculate the Energy Change

STEP 3

STEP 4

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

STEP 5

Let's **plug in** our values.We're given ni=4n_i = 4ni​=4 (the **initial level**) and nf=1n_f = 1nf​=1 (the **final level**).So, we have: ΔE=−hR(112−142) \Delta E = -hR \left( \frac{1}{1^2} - \frac{1}{4^2} \right) ΔE=−hR(121​−421​)

STEP 6

Now, let's **simplify** those fractions: ΔE=−hR(11−116) \Delta E = -hR \left( \frac{1}{1} - \frac{1}{16} \right) ΔE=−hR(11​−161​) ΔE=−hR(1−116) \Delta E = -hR \left( 1 - \frac{1}{16} \right) ΔE=−hR(1−161​)

STEP 7

To subtract the fractions, we need a **common denominator**, which is **16**: ΔE=−hR(1616−116) \Delta E = -hR \left( \frac{16}{16} - \frac{1}{16} \right) ΔE=−hR(1616​−161​)

STEP 8

Now, **subtract**: ΔE=−hR(16−116) \Delta E = -hR \left( \frac{16 - 1}{16} \right) ΔE=−hR(1616−1​) ΔE=−hR(1516) \Delta E = -hR \left( \frac{15}{16} \right) ΔE=−hR(1615​)ΔE=−1516hR \Delta E = -\frac{15}{16} hR ΔE=−1615​hR

STEP 9

The **energy change** is −1516hR-\frac{15}{16} hR−1615​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 ∣ΔE∣=1615​hR

STEP 10

The amount of energy associated with the transition is 1516hR\frac{15}{16} hR1615​hR.

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1. Rydberg Formula
2. Calculate the Energy Change

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

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

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

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

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

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

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