Question

Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum uncertainty in the electron's momentum

Answer 1

• Uncertainty in position (∆x) = 24 × 10⁻¹⁵ m
• Uncertainty in momentum (∆P) = ?
• Planck's constant (h) = 6.26 × 10⁻³⁴ Js

`\longrightarrow \: \: \sf\Delta x .\Delta p = (h)/(4\pi)`

`\longrightarrow \: \: \sf24 * {10}^( - 15) .\Delta p = \frac{6.26 * {10}^( - 34)} {4 * (22)/(7) }`

`\longrightarrow \: \: \sf24 * {10}^( - 15) .\Delta p = \frac{6.26 * {10}^( - 34)} { (88)/(7) }`

`\longrightarrow \: \: \sf24 * {10}^( - 15) .\Delta p = \frac{6.26 * {10}^( - 34) * 7} { 8 }`

`\longrightarrow \: \: \sf\Delta p = \frac{43.82 * {10}^( - 34) } { 8 * 24 * {10}^( - 15) }`

`\longrightarrow \: \: \sf\Delta p = \frac{43.82 * {10}^( - 34) } { 192 * {10}^( - 15) }`

`\longrightarrow \: \: \sf\Delta p = \frac{43.82 * {10}^( - 34) * {10}^(15) } { 192}`

`\longrightarrow \: \: \sf\Delta p = \frac{43.82 * {10}^( -19) } { 192}`

`\longrightarrow \: \: \sf\Delta p = \frac{4382 * {10}^( - 2) * {10}^( -19) } { 192}`

`\longrightarrow \: \: \sf\Delta p = \frac{4382 * {10}^( - 21) } { 192}`

`\longrightarrow \: \: \sf\Delta p = 22.822* {10}^( - 21)`

`\longrightarrow \: \: \sf\Delta p = 2.2822 * {10}^(1) * {10}^( - 21)`

`\longrightarrow \: \: \underline{ \boxed{ \red{ \bf\Delta p = 2.2822 * {10}^( - 20) \: kg/ms}}}`