Question

What is the minimum amount of time required to measure the kinetic energy of an electron with a speed of (100+_1) ms-1

Answer 1

we have to find the minimum amount of time required to measure the kinetic energy of an electron with a speed of (100 ± 1) m/s.

solution : from Heisenberg's uncertainty principle,

∆E ∆t ≤ h/4π ......(1)

uncertainty in kinetic energy of electron, ∆E = 1/2 m∆v²

here mass of electron, m = 9.1 × 10¯³¹ kg

uncertainty in velocity, ∆v = 1 m/s

so ∆E = 1/2 × 9.1 × 10¯³¹ × 1² = 4.55 × 10¯³¹ J

hare h = 6.63 × 10¯³⁴ Js

now from equation (1),

4.55 × 10¯³¹ × ∆t ≤ (6.63 × 10¯³⁴)/(4 × 3.14)

⇒∆t ≤ (6.63 × 10¯³)/(4 × 3.14 × 4.55)

⇒∆t ≤ 0.116 × 10¯³ sec

Therefore the minimum amount of time would be 0.116 ms.