Energy of an atom in an excited state

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SUMMARY

The discussion revolves around calculating the energy of an atom in an excited state, specifically when it emits a photon with a wavelength of 330 nm. The participant correctly applies the wave formula and the photon energy formula, E_{photon} = hf, to derive the energy of the emitted photon as 6.03 x 10^{-19} J. By combining this with the energy of the last energy level, initially stated as -4.8 x 10^{-19} J, the participant calculates the excited state energy to be 1.23 x 10^{-19} J. However, the correct energy of the last level is revealed to be -7.7 x 10^{-19} J, leading to the final energy of the excited state being -0.17 x 10^{-18} J.

PREREQUISITES
  • Understanding of photon energy calculations using E_{photon} = hf
  • Familiarity with the wave equation c = fλ
  • Knowledge of Bohr's model of the atom, specifically E_{state} = -B/n^2
  • Basic algebra for manipulating equations and solving for unknowns
NEXT STEPS
  • Study the implications of Bohr's model on atomic energy levels
  • Learn about the relationship between wavelength and energy in quantum mechanics
  • Explore advanced photon emission processes in different atomic structures
  • Investigate common errors in energy level calculations and their resolutions
USEFUL FOR

Students studying quantum mechanics, physics educators, and anyone interested in atomic energy transitions and photon emissions.

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Homework Statement


An atom which is in an excited state, can emit a photon with the wavelength of 330 nm at during the movement between two "energy niveaus". The last niveau has the energy -4.8 \cdot 10^{-19}J. What is the energy of the atom in the excited state?

Homework Equations



Not relevant for my conseptual question, but for solving the problem

the wave formula:c=f \lambda and the formula for the energy of the photonE_{photon} =hf where h=6.63 \cdot 10^{-34}

Also Bohrs formula: E_{state}=-\frac{B}{n^2} where n is the number of the state

The Attempt at a Solution



I just wonder if I have understood the problem right...

I can use the wavelength to find the energy of the emitted photon . Then I combine the formula for the energy of the photon with Bohrs formula to find the energy of the atom in the excited state, right?

The energy in the excited state will be the energy in the last niveau plus the energy of the emitted photon, won't it?

These are my calculations:

The frequency for the emitted photon: c=f\lambda gives f=\frac{c}{\lambda}=\frac{3.00 \cdot 10^8 m/sec}{330 \cdot 10^{-9} m}= 9.09 \cdot 10^{14} s^{-1}= 9.09 \cdot 10^{14} Hz

The energy of the photon: E_{photon} = 6.63 \cdot 10^{-34} Js \cdot 9.09 s^{-1} = 6.03 \cdot 10^{-19}

The energy of the atom in the excited state: E_{excited}=-4.8 \cdot 10^{-19} J + 6.03 \cdot 10^{-19}=1.23 \cdot 10^{-19}

However the answer shall be, according to my book: -0,17 \cdot 10^{-18} J (I also know I shouldn't have got a positive answer)

Can anybody see what has gone wrong?
 
Last edited:
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There was an error in my book.

The energy of the last niveau should be -7.7 \cdot 10^{-19} in the problem text.

Thus my problem is now solved :smile:
 

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