Question

An excited state of a particular atom has a mean lifetime of 6.2×10−10 s , which we may take as the uncertainty δt.

Answer 1

According to Heisenberg's Uncertainty Principle,
ΔE×Δt ≥ h/(4π)
where h = planck's constant = 6.63 ×
`10^(-34)` J.s

Hence, ΔE ≥
`(h)/(4.pi.delta(t))` =
`(6.63X 10^(-34) )/(4X3.142X6.2X 10^(-10))`
∴, ΔE ≥ 8.5086 X
`10^(-26)` J