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 calculate the average lifetime of the first excited state of a hydrogen atom, we need to use the uncertainty principle and the relationship between energy and time. According to the uncertainty principle, there is a fundamental uncertainty in the measurement of both energy and time.
The uncertainty principle states: ΔE * Δt ≥ ħ/2
Where ΔE is the uncertainty in energy, Δt is the uncertainty in time, and ħ (pronounced "h-bar") is the reduced Planck's constant.
To calculate the average lifetime, we need to find the uncertainty in energy (ΔE). Since the energy is given as -0.34 eV ± 0.0003 eV, the uncertainty in energy is ± 0.0003 eV.
Now, we can rearrange the uncertainty principle equation to solve for the uncertainty in time (Δt):
Δt ≥ ħ / (2 * ΔE)
Where ħ has a value of approximately 6.63 × 10^-34 J·s.
To convert the energy uncertainty from eV to Joules, we need to use the conversion factor: 1 eV = 1.602 × 10^-19 J.
Now let's plug in the values to calculate Δt:
Δt ≥ (6.63 × 10^-34 J·s) / (2 * (0.0003 eV * 1.602 × 10^-19 J/eV))
Simplifying the equation, we get:
Δt ≥ (6.63 × 10^-34 J·s) / (2 * 0.0003 * 1.602 × 10^-19 J)
Δt ≥ (6.63 × 10^-34 J·s) / (0.0003 * 1.602 × 10^-19 J)
Δt ≥ (6.63 × 10^-34 J·s) / (4.806 × 10^-22 J)
Δt ≥ 1.376 × 10^-12 s
Therefore, the average lifetime of the first excited state of a hydrogen atom is approximately 1.376 × 10^-12 seconds.