Question

(a) How much time is needed to measure the kinetic energy of an electron whose speed is \( 10.0 \mathrm{~m} / \mathrm{s} \) with an uncertainty of no more than 0.100 percent? How far will the electron have travelled in this period of time? (b) Make the same calculations as in (a) for a \( \mathbf{1 . 0 0 - g} \) insect whose speed is the same. What do these sets of figures indicate? 14. How much time is needed to measure the kinetic energy of an electron whose speed is \( 10.0 \mathrm{~m} / \mathrm{s} \) with an uncertainty of no more than 0.100 percent? How far will the electron have travelled in this period of time? (b) Make the same calculations for a \( 1.00-\mathrm{g} \) insect whose speed is the same. What do these sets of figures indicate? (Beiser, Ex. 34, pg. 118) Solution 3-34: (a) The uncertainty in the electron's energy is \( \Delta E=1.00 \times 10^{-3} E= \) | \( 10^{-3}\left(\mathrm{mv}^{2} / 2\right) \), and so the minimum time needed to measure the energy is \[ \Delta t>\frac{h}{4 \pi \Delta E}=\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s}\right)}{4 \pi\left(10^{-3}\right)\left(9.1095 \times 10^{-31} \mathrm{~kg}\right)(10 \mathrm{~m} / \mathrm{s})^{2} / 2}=1.16 \times 10^{-3} \mathrm{~s} \]

          (a) How much time is needed to measure the kinetic energy of an electron whose speed is \( 10.0 \mathrm{~m} / \mathrm{s} \) with an uncertainty of no more than 0.100 percent? How far will the electron have travelled in this period of time?
(b) Make the same calculations as in (a) for a \( \mathbf{1 . 0 0 - g} \) insect whose speed is the same. What do these sets of figures indicate?
14. How much time is needed to measure the kinetic energy of an electron whose speed is \( 10.0 \mathrm{~m} / \mathrm{s} \) with an uncertainty of no more than 0.100 percent? How far will the electron have travelled in this period of time? (b) Make the same calculations for a \( 1.00-\mathrm{g} \) insect whose speed is the same. What do these sets of figures indicate? (Beiser, Ex. 34, pg. 118)
Solution

3-34: (a) The uncertainty in the electron's energy is \( \Delta E=1.00 \times 10^{-3} E= \) | \( 10^{-3}\left(\mathrm{mv}^{2} / 2\right) \), and so the minimum time needed to measure the energy is
\[
\Delta t>\frac{h}{4 \pi \Delta E}=\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s}\right)}{4 \pi\left(10^{-3}\right)\left(9.1095 \times 10^{-31} \mathrm{~kg}\right)(10 \mathrm{~m} / \mathrm{s})^{2} / 2}=1.16 \times 10^{-3} \mathrm{~s}
\]

(a) How much time is needed to measure the kinetic energy of an electron whose speed is 10.0 m / s with an uncertainty of no more than 0.100 percent? How far will the electron have travelled in this period of time?
(b) Make the same calculations as in (a) for a 1 . 0 0 - 𝐠 insect whose speed is the same. What do these sets of figures indicate?
14. How much time is needed to measure the kinetic energy of an electron whose speed is 10.0 m / s with an uncertainty of no more than 0.100 percent? How far will the electron have travelled in this period of time? (b) Make the same calculations for a 1.00-g insect whose speed is the same. What do these sets of figures indicate? (Beiser, Ex. 34, pg. 118)
Solution

3-34: (a) The uncertainty in the electron's energy is Δ E=1.00 × 10^-3 E= | 10^-3(mv^2 / 2), and so the minimum time needed to measure the energy is

Δ t>(h)/(4 πΔ E)=((6.626 × 10^-34 J· s))/(4 π(10^-3)(9.1095 × 10^-31 kg)(10 m / s)^2 / 2)=1.16 × 10^-3 s

Added by Aaron M.

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