Question

Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter \( 10^{-15} \, \text{m} \) is \( \dots \dots  \times 10^9 \, \text{ms}^{-1} \) (nearest integer). \[ \text{[Given: mass of electron} = 9.1 \times 10^{-31} \, \text{kg, Planck's constant (} h \text{)} = 6.626 \times 10^{-34} \, \text{Js]} \] \[ \text{(Value of } \pi = 3.14) \]

Answer 1

Correct Answer: 58

From Heisenberg's uncertainty principle:
\[ \Delta x \cdot m_e \cdot \Delta v \geq \frac{h}{4\pi} \]
Here:
\[ \Delta x = 10^{-15} \, \text{m}, \quad m_e = 9.1 \times 10^{-31} \, \text{kg}, \quad h = 6.626 \times 10^{-34} \, \text{Js}. \]
Rearranging for the uncertainty in velocity ($\Delta v$):
\[ \Delta v \geq \frac{h}{4\pi \cdot \Delta x \cdot m_e} \]
Substitute the values:
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{4 \cdot 3.14 \cdot (10^{-15}) \cdot (9.1 \times 10^{-31})} \]
Simplify the denominator:
\[ 4 \cdot 3.14 \cdot 10^{-15} \cdot 9.1 \times 10^{-31} = 1.143 \times 10^{-44} \]
Substitute back:
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{1.143 \times 10^{-44}} = 5.8 \times 10^{10} \, \text{ms}^{-1} \]
Uncertainty in velocity:
\[ \Delta v = 58 \times 10^9 \, \text{ms}^{-1} \]
Final Answer: 58.