Question

The lifetime of an excited state of an atom is $3\times {10}^3$ s. What will be the minimum uncertainity in the Energy?

• A. $1.76\times {10}^{-35}$ J
• B. $1.76\times {10}^{-34}$ J
• C. $1.76\times {10}^{-33}$ J
• D. $1.76\times {10}^{-38}$ J

Answer 1

The correct option is D $1.76\times {10}^{-38}$ J

We know Energy-time uncertainity principle
ΔE. Δt ≥ $\frac{h}{4\pi }$
Assuming minimum error in lifetime measurement = $3\times {10}^3$
$\therefore \Delta E\ge \frac{h}{4\pi \Delta t}$ = $\frac{6.626\times {10}^{-34}}{4\times 3.14\times 3\times {10}^3}$
On solving, ΔE = $1.76\times {10}^{-38}$ J