Question

the uncertanity in energy of photon which is emmited from an atom radiating for 10^_8 second is ?

Answer 1

We know that according to Heisenberg's Uncertainity Principle : the product of uncertainity in energy of particle and uncertainity in time to assess that energy is greater than or equal to h/4π

⇒ Δ(energy) × Δ(time) ≥ h/4π

Given Δ(time) = 10⁻⁸ and where h is planks constant = 6.625 × 10⁻³⁴ J-sec

substituting in the above equation we get:

⇒ Δ(energy) × 10⁻⁸ ≥ h/4π

⇒ Δ(energy) ≥ (6.625 × 10⁻³⁴)/(4π × 10⁻⁸)

⇒ Δ(energy) ≥ 5.27 × 10⁻²⁷Joules

⇒ The uncertainity in the energy of photon when measured with an uncertainity in time of 10⁻⁸ sec is greater than or equal to 5.27 × 10⁻²⁷Joules