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). \[ \text{[Given: mass of electron} = 9.1 \times 10^{-31} \, \text{kg, Planck's constant (} h \text{)} = 6.626 \times 10^{-34} \, \text{Js]} \] \[ \text{(Value of } \pi = 3.14) \]
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). \[ \text{[Given: mass of electron} = 9.1 \times 10^{-31} \, \text{kg, Planck's constant (} h \text{)} = 6.626 \times 10^{-34} \, \text{Js]} \] \[ \text{(Value of } \pi = 3.14) \]
Correct Answer: 58
Approach Solution - 1
To find the uncertainty in the velocity of an electron within an atomic nucleus using Heisenberg's uncertainty principle, we have: \[\Delta x \cdot \Delta p \geq \frac{h}{4\pi}\] where \(\Delta x\) is the uncertainty in position and \(\Delta p\) is the uncertainty in momentum. Given \(\Delta x = 10^{-15} \, \text{m}\), we want to find \(\Delta v\), the uncertainty in velocity. \(\Delta p = m \cdot \Delta v\), where \(m\) is the mass of the electron.
Plug in the known values: \[\Delta x \cdot m \cdot \Delta v \geq \frac{h}{4\pi}\] \[10^{-15} \cdot 9.1 \times 10^{-31} \cdot \Delta v \geq \frac{6.626 \times 10^{-34}}{4 \times 3.14}\]
Calculate the right-hand side: \[\frac{6.626 \times 10^{-34}}{12.56} \approx 5.274 \times 10^{-35}\]
Substitute and solve for \(\Delta v\): \[10^{-15} \cdot 9.1 \times 10^{-31} \cdot \Delta v \geq 5.274 \times 10^{-35}\]
Rearrange to solve for \(\Delta v\): \[\Delta v \geq \frac{5.274 \times 10^{-35}}{9.1 \times 10^{-46}}\] \[\Delta v \geq 5.796 \times 10^{10} \, \text{m/s}\]
Express \(\Delta v\) in scientific notation to the nearest integer: \[5.796 \times 10^{10} \approx 58 \times 10^9 \, \text{m/s}\]
Thus, the uncertainty in the velocity of the electron is \(58 \times 10^9 \, \text{ms}^{-1}\). Verifying this value falls within the expected range, we conclude the calculation aligns perfectly with the expected outcome.
Approach Solution -2
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.
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