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QuestionCalculate the wavelength for a hydrogen atom transition from (n=2)(n=2) to (n=4)(n=4) using λ=hcΔE\lambda=\frac{h c}{\Delta E}. What color is it?

Studdy Solution

STEP 1

Assumptions1. We are dealing with a hydrogen atom. The initial state of the atom is n=n=
3. The final state of the atom is n=4n=4
4. The formula for the energy difference between two states in a hydrogen atom is \Delta =_{final} -_{initial} = -13.6 \, eV \, \left( \frac{1}{n_{final}^} - \frac{1}{n_{initial}^} \right)5. The formula for the wavelength of light absorbed is λ=hcΔ\lambda = \frac{hc}{\Delta}, where hh is Planck's constant, cc is the speed of light, and Δ\Delta is the energy difference between the two states6. The energy difference Δ\Delta must be converted from electron volts (eV) to joules (J) for the formula to work, using the conversion 1 eV=1.602×10−19 J1 \, eV =1.602 \times10^{-19} \, J
7. The color of the light corresponds to the wavelength, with violet light having the shortest wavelength and red light having the longest

STEP 2

First, we need to calculate the energy difference between the two states. We can do this using the formula for Δ\Delta.
Δ=−13.6 eV (1nfinal2−1ninitial2)\Delta = -13.6 \, eV \, \left( \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} \right)

STEP 3

Now, plug in the given values for ninitialn_{initial} and nfinaln_{final} to calculate Δ\Delta.
Δ=−13.6 eV (12−122)\Delta = -13.6 \, eV \, \left( \frac{1}{^2} - \frac{1}{2^2} \right)

STEP 4

Calculate the energy difference Δ\Delta.
Δ=−13.6 eV (116−14)=1.71 eV\Delta = -13.6 \, eV \, \left( \frac{1}{16} - \frac{1}{4} \right) =1.71 \, eV

STEP 5

Convert the energy difference from electron volts (eV) to joules (J) using the conversion 1 eV=1.602×10−19 J1 \, eV =1.602 \times10^{-19} \, J.
Δ=1.71 eV×1.602×10−19 J/eV\Delta =1.71 \, eV \times1.602 \times10^{-19} \, J/eV

STEP 6

Calculate the energy difference in joules.
Δ=1.71 eV×1.602×10−19 J/eV=2.74×10−19 J\Delta =1.71 \, eV \times1.602 \times10^{-19} \, J/eV =2.74 \times10^{-19} \, J

STEP 7

Now that we have the energy difference, we can calculate the wavelength of light absorbed using the formula λ=hcΔ\lambda = \frac{hc}{\Delta}, where hh is Planck's constant (6.626×10−34 J⋅s6.626 \times10^{-34} \, J \cdot s) and cc is the speed of light (3.00 \times10^ \, m/s).
λ=hcΔ\lambda = \frac{hc}{\Delta}

STEP 8

Plug in the values for hh, cc, and Δ\Delta to calculate the wavelength.
λ=6.626×10−34 J⋅s×3.00×108 m/s2.74×10−19 J\lambda = \frac{6.626 \times10^{-34} \, J \cdot s \times3.00 \times10^8 \, m/s}{2.74 \times10^{-19} \, J}

STEP 9

Calculate the wavelength of light absorbed.
λ=6.626×−34 J⋅s×3.00×8 m/s2.74×−19 J=7.25×−7 m\lambda = \frac{6.626 \times^{-34} \, J \cdot s \times3.00 \times^8 \, m/s}{2.74 \times^{-19} \, J} =7.25 \times^{-7} \, m

STEP 10

The wavelength of light absorbed corresponds to the color of the light. A wavelength of 7.25×10−7 m7.25 \times10^{-7} \, m falls in the red region of the visible spectrum.
The wavelength of light that must be absorbed by a hydrogen atom to go from the state (n=2)(n=2) to the excited state (n=4)(n=4) is 7.25×10−7 m7.25 \times10^{-7} \, m, which corresponds to red light.

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