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The discussion focuses on calculating the energy uncertainty of the first excited state of a hydrogen atom, given its lifetime of approximately 10-8 seconds. Using the formula dE = h/2π, the energy uncertainty is determined to be 7 x 10-8 eV. This value is significantly smaller than the energy of the first excited state, which is calculated to be 3.4 eV. The calculations utilize Planck's constant (h = 6.63 x 10-34 J·s) and the conversion factor from joules to electron volts (1.6 x 10-19 J/eV).

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Please I want a solution to this question??

This question is Modern Physics 2

Q The lifetimes of the levels in a hydrogen atom are of the order of 10-8 s. Find the energy uncertainty of the first excited state and compare it with the energy of the state

The final answer is 7*10^-8ev

Thank you
 
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momd said:
This question is Modern Physics 2

Q The lifetimes of the levels in a hydrogen atom are of the order of 10-8 s. Find the energy uncertainty of the first excited state and compare it with the energy of the state

The final answer is 7*10^-8ev

Thank you
dEdT = h/2Pi
dE = (6.63 x 10^-34)/(2Pi * 10^-8) J
dE = [(6.63 x 10^-34)/(2Pi * 10^-8 * 1.6 x10^-19)] eV
dE = 6.6 x 10^-8 eV
dE = 7 x 10^-8 eV (1 Sig Fig)

The energy of the first excited state is 13.6/4 = 3.4 eV, so its energy uncertainty is tiny compared in comparison.
 


Physics Enemy said:
dEdT = h/2Pi
dE = (6.63 x 10^-34)/(2Pi * 10^-8) J
dE = [(6.63 x 10^-34)/(2Pi * 10^-8 * 1.6 x10^-19)] eV
dE = 6.6 x 10^-8 eV
dE = 7 x 10^-8 eV (1 Sig Fig)

The energy of the first excited state is 13.6/4 = 3.4 eV, so its energy uncertainty is tiny compared in comparison.


Thank you very much
 

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