Please I want a solution to this question?

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The discussion revolves around calculating the energy uncertainty of the first excited state of a hydrogen atom, given its lifetime of approximately 10^-8 seconds. The derived energy uncertainty is approximately 7 x 10^-8 eV, which is significantly smaller than the energy of the first excited state, calculated to be 3.4 eV. The calculations utilize the formula dE = h/(2πΔt) to arrive at the uncertainty value. This comparison highlights that the energy uncertainty is negligible relative to the energy of the state. The thread concludes with gratitude for assistance in solving the problem.
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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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