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Chemistry Question on Atomic Structure

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 m10^{-15} \, \text{m} is …⋯×109 ms−1\dots \dots \times 10^9 \, \text{ms}^{-1} (nearest integer). [Given: mass of electron=9.1×10−31 kg, Planck’s constant (h)=6.626×10−34 Js]\text{[Given: mass of electron} = 9.1 \times 10^{-31} \, \text{kg, Planck's constant (} h \text{)} = 6.626 \times 10^{-34} \, \text{Js]} (Value of π=3.14)\text{(Value of } \pi = 3.14)

Answer

From Heisenberg's uncertainty principle:
Δx⋅me⋅Δv≥h4π\Delta x \cdot m_e \cdot \Delta v \geq \frac{h}{4\pi}
Here:
Δx=10−15 m,me=9.1×10−31 kg,h=6.626×10−34 Js.\Delta x = 10^{-15} \, \text{m}, \quad m_e = 9.1 \times 10^{-31} \, \text{kg}, \quad h = 6.626 \times 10^{-34} \, \text{Js}.
Rearranging for the uncertainty in velocity (Δv\Delta v):
Δv≥h4π⋅Δx⋅me\Delta v \geq \frac{h}{4\pi \cdot \Delta x \cdot m_e}
Substitute the values:
Δv≥6.626×10−344⋅3.14⋅(10−15)⋅(9.1×10−31)\Delta v \geq \frac{6.626 \times 10^{-34}}{4 \cdot 3.14 \cdot (10^{-15}) \cdot (9.1 \times 10^{-31})}
Simplify the denominator:
4⋅3.14⋅10−15⋅9.1×10−31=1.143×10−444 \cdot 3.14 \cdot 10^{-15} \cdot 9.1 \times 10^{-31} = 1.143 \times 10^{-44}
Substitute back:
Δv≥6.626×10−341.143×10−44=5.8×1010 ms−1\Delta v \geq \frac{6.626 \times 10^{-34}}{1.143 \times 10^{-44}} = 5.8 \times 10^{10} \, \text{ms}^{-1}
Uncertainty in velocity:
Δv=58×109 ms−1\Delta v = 58 \times 10^9 \, \text{ms}^{-1}
Final Answer: 58.

Explanation

Solution

From Heisenberg's uncertainty principle:
Δx⋅me⋅Δv≥h4π\Delta x \cdot m_e \cdot \Delta v \geq \frac{h}{4\pi}
Here:
Δx=10−15 m,me=9.1×10−31 kg,h=6.626×10−34 Js.\Delta x = 10^{-15} \, \text{m}, \quad m_e = 9.1 \times 10^{-31} \, \text{kg}, \quad h = 6.626 \times 10^{-34} \, \text{Js}.
Rearranging for the uncertainty in velocity (Δv\Delta v):
Δv≥h4π⋅Δx⋅me\Delta v \geq \frac{h}{4\pi \cdot \Delta x \cdot m_e}
Substitute the values:
Δv≥6.626×10−344⋅3.14⋅(10−15)⋅(9.1×10−31)\Delta v \geq \frac{6.626 \times 10^{-34}}{4 \cdot 3.14 \cdot (10^{-15}) \cdot (9.1 \times 10^{-31})}
Simplify the denominator:
4⋅3.14⋅10−15⋅9.1×10−31=1.143×10−444 \cdot 3.14 \cdot 10^{-15} \cdot 9.1 \times 10^{-31} = 1.143 \times 10^{-44}
Substitute back:
Δv≥6.626×10−341.143×10−44=5.8×1010 ms−1\Delta v \geq \frac{6.626 \times 10^{-34}}{1.143 \times 10^{-44}} = 5.8 \times 10^{10} \, \text{ms}^{-1}
Uncertainty in velocity:
Δv=58×109 ms−1\Delta v = 58 \times 10^9 \, \text{ms}^{-1}
Final Answer: 58.