Question

Consider an electron with a mass of 9.11 x 1051 kg and a 100.0 g tennis ball that are both moving with a velocity of 70.0 m s1. (a) Calculate the momentum of the electron (p mv). (b) Calculate the momentum of the tennis ball (c) What is the uncertainty in the position of the electron (Ax) if the uncertainty in its momentum (Ap) is equal to 1.0 % of p for the electron? (d) What is the uncertainty in the position of the tennis ball (Ax) if the uncertainty in momentum (Ap) is equal to 0.1% of p for the tennis ball? (e) Comment on how the uncertainty in position (Ax) compares to the overall size in each case

Answer 1

Answer:

(a) 6.38 × 10⁻²⁹ kg·m·s⁻¹; (b) 7.00 kg·m·s⁻¹; (c) 82.7 µm; (d) 7.53 × 10⁻³⁴  m;

(e) Δx ∝ 1/m

Explanation:

(a) Momentum of electron

p = mv = 9.11  × 10⁻³¹ kg × 70.0 m·s⁻¹ = 6.38 × 10⁻²⁹ kg·m·s⁻¹

(b) Momentum of tennis ball

p = mv = 0.1000 kg × 70.0 m·s⁻¹ = 7.00 kg·m·s⁻¹

(c) Δx for electron

Δp = 0.010p = 0.010 × 6.38 × 10⁻²⁹ kg·m·s⁻¹ = 6.38 × 10⁻³¹ kg·m·s⁻¹

$\begin{array}{rcl}\Delta x \Delta p & \geq & \dfrac{h}{4 \pi}\\\\\Delta x \times 6.38 \times 10^{-31} \text{ kg \cdot m \cdot s ^{-1} } & \geq & \dfrac{6.626 \times 10^{-34} \text{ kg \cdot m ^{2} s}^{-1}}{4 \pi}\\\\\Delta x \times 6.38 \times 10^{-31} & \geq & 5.273 \times 10^{-35} \text{ m}\\\Delta x & \geq & \dfrac{5.273 \times 10^{-35} \text{ m}}{6.38 \times 10^{-31}}\\\\ & \geq&8.27 \times10^{-5} \text{ m}\\ &\geq&\textbf{82.7 \mu m}\\\end{array}$

(d) Δx for tennis ball

Δp = 0.010p = 0.010 × 7.00 kg·m·s⁻¹ = 0.0700 kg·m·s⁻¹

$\begin{array}{rcl}\Delta x \times 0.0700 & \geq & 5.273 \times 10^{-35} \text{ m}\\\Delta x & \geq & \dfrac{5.273 \times 10^{-35} \text{ m}}{0.0700}\\\\ &\geq& \textbf{7.53 \mathbf{\times 10^{-34}} m}\\\end{array}$

(e) Relative uncertainty

Both particles are travelling at the same speed, so,

ΔxΔp = Δx × mv = mvΔx = constant

v is constant, so  

Δx ∝ 1/m

Thus, the larger the mass of an object, the smaller the uncertainty in its velocity.