I Review | Constants | Periodic Table involving transition of the electron from two energy states ny to ng is given as ΔΕ=-(2.18 x 10-18 J) x (-) Niels Bohr is credited with deriving this form of the Rydberg equation, which is sometimes called Bohr's equation Part B What is the change in energy, AE, in kilojoules per mole of hydrogen atoms for an electron transition from n-4 to n= 2? Express your answer in kilojoules per mole to three significant figures. > View Available Hint(s) x" X•10" AE = - 301 kJ/mol

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# Chapter 8 Problem Set

## Derivations from the Rydberg Equation

The Rydberg equation expresses the wavelength, λ, of emitted light based on the initial and final energy states (n₁ and n₂) of an electron in a hydrogen atom:

\[
\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\]

where 
\[ 
R_H = 1.097 \times 10^7 \, \text{m}^{-1} = 1.097 \times 10^{-2} \, \text{nm}^{-1} 
\]    

You may also see this equation written as:

\[
\frac{1}{\lambda} = -R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right)
\]

Since

\[
\frac{1}{n_1^2} - \frac{1}{n_2^2} = -\left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right)
\]

The two formulas are equivalent and sometimes used interchangeably. It can help to remember that when light is emitted, E is negative. When light is absorbed, E is positive.

### Energy Change

The change in energy, ΔE, involving the transition of the electron from two energy states n₀ to n₁, is given as:

\[
\Delta E = \left(-2.18 \times 10^{-18} \, \text{J} \right) \times \left( \frac{1}{n_1^2} - \frac{1}{n_0^2} \right)
\]

Niels Bohr is credited with deriving this form of the Rydberg equation, which is sometimes called Bohr's equation.

### Part B

What is the change in energy, ΔE, in kilojoules per mole of hydrogen atoms for an electron transition from n = 4 to n = 2? Express your answer in kilojoules per mole to three significant figures.

**Answer Input:**

ΔE = \[ \boxed{301} \, \text{kJ/mol
Transcribed Image Text:# Chapter 8 Problem Set ## Derivations from the Rydberg Equation The Rydberg equation expresses the wavelength, λ, of emitted light based on the initial and final energy states (n₁ and n₂) of an electron in a hydrogen atom: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \[ R_H = 1.097 \times 10^7 \, \text{m}^{-1} = 1.097 \times 10^{-2} \, \text{nm}^{-1} \] You may also see this equation written as: \[ \frac{1}{\lambda} = -R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \] Since \[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = -\left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \] The two formulas are equivalent and sometimes used interchangeably. It can help to remember that when light is emitted, E is negative. When light is absorbed, E is positive. ### Energy Change The change in energy, ΔE, involving the transition of the electron from two energy states n₀ to n₁, is given as: \[ \Delta E = \left(-2.18 \times 10^{-18} \, \text{J} \right) \times \left( \frac{1}{n_1^2} - \frac{1}{n_0^2} \right) \] Niels Bohr is credited with deriving this form of the Rydberg equation, which is sometimes called Bohr's equation. ### Part B What is the change in energy, ΔE, in kilojoules per mole of hydrogen atoms for an electron transition from n = 4 to n = 2? Express your answer in kilojoules per mole to three significant figures. **Answer Input:** ΔE = \[ \boxed{301} \, \text{kJ/mol
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