) The energy of the first excited state of a hydrogen atom is -0.34 eV ± 0.0003 eV. What is the average lifetime of for this state?
To find the average lifetime of the first excited state of a hydrogen atom, you need to use the energy-time uncertainty principle. According to this principle, the uncertainty in energy (ΔE) and the uncertainty in time (Δt) are related by the equation ΔE Δt ≥ ħ/2, where ħ is the reduced Planck's constant.
In this case, the uncertainty in energy is given as ±0.0003 eV. Since this value represents the uncertainty, we will take the positive value, 0.0003 eV, for simplicity.
Now, we can substitute the values into the equation and solve for the average lifetime (Δt):
ΔE Δt ≥ ħ/2
(0.0003 eV) Δt ≥ (ħ/2)
Δt ≥ (ħ/2) / (0.0003 eV)
To perform this calculation, we need to convert ħ (reduced Planck's constant) from its SI unit to electron volts (eV). The value of ħ is approximately 4.136 x 10^(-15) eV s.
Δt ≥ (4.136 x 10^(-15) eV s / 2) / (0.0003 eV)
Simplifying further:
Δt ≥ (2.068 x 10^(-15) eV s) / (0.0003 eV)
Δt ≥ 6.893 x 10^(-12) s
Therefore, the average lifetime of the first excited state of a hydrogen atom is approximately 6.893 x 10^(-12) seconds.