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
[tex]\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}[/tex]
(d) Δx for tennis ball
Δp = 0.010p = 0.010 × 7.00 kg·m·s⁻¹ = 0.0700 kg·m·s⁻¹
[tex]\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}[/tex]
(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.
