Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter 10⁻¹⁵ m is ________ × 10⁹ ms⁻¹ (nearest integer)
[Given : mass of electron =9.1 × 10⁻³¹ kg, Plank's constant (h)=6.626 × 10⁻³⁴ Js] (Value of π=3.14)

Solution

Chemistry ma’am · 4 steps

Take the nucleus diameter as Δ x, then Δ v = h/(4π m Δ x) does all the work.

1. Set up Heisenberg. The electron must lie somewhere inside the nucleus, so its position uncertainty is the full diameter: Δ x = 10⁻¹⁵ m. Write Δ p = m Δ v and take the minimum uncertainty.

   Δ x · m Δ v = h/4π ⇒ Δ v = h/(4π m Δ x)

2. Denominator first. 4π m Δ x = 4 × 3.14 × 9.1 × 10⁻³¹ × 10⁻¹⁵ = 1.143 × 10⁻⁴⁴ kg m.
3. Divide. Δ v = (6.626 × 10⁻³⁴)/(1.143 × 10⁻⁴⁴) = 5.797 × 10¹⁰ m s⁻¹ = 57.97 × 10⁹ m s⁻¹.
4. So. To the nearest integer, the uncertainty in velocity is 58 × 10⁹ m s⁻¹. That is far faster than light, which is why an electron cannot live inside a nucleus.

The common slip: 116. Using the radius 0.5 × 10⁻¹⁵ m instead of the diameter as Δ x gives 116, not 58.

The idea to keep. In confinement questions, Δ x is the whole size of the region the particle can be in. Use Δ v = h/(4π mΔ x) and watch the powers of ten carefully.

Answer from the NTA final answer key. Solution by PrepWiser.