The error in the measurement of lifetime of an atom is 2 × 10^-2 sec. What is the minimum uncertainty in its energy in eV?

Concept:

Heisenberg's uncertainty principle states that for particles exhibiting both particle and wave nature, it will not be possible to accurately determine both the position and velocity at the same time.

Heisenberg’s principle is applicable to all matter waves. The measurement error of any two conjugate properties, whose dimensions happen to be joule sec, like position-momentum, time-energy will be guided by the Heisenberg’s value.

Heisenberg’s uncertainty principle can be considered as a very precise mathematical statement that describes the nature of quantum systems. As such, we often consider two common equations related to the uncertainty principle.

They are,

\({\rm{Δ }}x\;.\;{\rm{Δ }}p \ge \frac{h}{{4\pi }}\)

\({\rm{Δ }}E\;.\;{\rm{Δ }}t \ge \frac{h}{{4\pi }}\)

Calcultaion:

Given Δt = 2 × 10^-2 sec;

From the second equation,

\({\rm{Δ }}E \ge \frac{h}{{4\pi \left( {{\rm{Δ }}t} \right)}} = \frac{{6.63 × {{10}^{ - 34}}}}{{4 × \pi × 2 × {{10}^{ - 2}}}} = 2.638 × {10^{ - 33}}\;J\)

Therefore, minimum uncertainty in energy is 

ΔE = 2.638 × 10^-33 J;

Converting in eV,

ΔE = (2.638 × 10^-33)/(1.6 × 10^-19) eV = 1.6 × 10^-14 eV;