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zmey [24]
3 years ago
12

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward

s a gold nucleus and its path was substantially deflected by the Coulomb interaction. If the initial kinetic energy of the doubly charged alpha nucleus was 5.28 MeV, how close to the gold nucleus (79 protons) could it come before being turned around
Physics
1 answer:
BaLLatris [955]3 years ago
3 0

Answer:

= 4.3 × 10 ⁻¹⁴ m

Explanation:

The alpha particle will be deflected when its kinetic energy is equal to the potential energy

Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C

Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷C

Kinetic energy of  the alpha particle = 5.28 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV)

=  8.459 × 10⁻¹³

k electrostatic force constant = 9 × 10⁹ N.m²/c²

Kinetic energy = potential energy =   k q₁q₂ / r where r is the closest distance the alpha particle got to the gold nucleus

r = (  9 × 10⁹ N.m²/c² × 3.2 × 10⁻¹⁹ C × 1.264 × 10⁻¹⁷C) / 8.459 × 10⁻¹³

= 4.3 × 10 ⁻¹⁴ m

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In the lab downstairs physics majors use a rotating mirror to measure the speed of light within a few percent of the actual valu
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Angular velocity is the ratio of the angle turned to the time taken.

The kinematic equation for angular velocity are presented as follows;

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10 months ago
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21) Acceleration from D to E: 1 m/s^2

22) The acceleration of the bus from D to E is 1 m/s^2

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Answer:

Hi

Final temperature = 250.11 °C

Final volume = 0,1 m3.

Process work = 0

Explanation:

The specific volume in the initial state is: v = 0.1m3/2 kg = 0.05 m3/kg.

This volume is located between the volumes as saturated liquid and saturated steam at 20 °C. For this reason the water is initially in a liquid vapor mixture. As the piston was blocked the volume remains constant and the process is isometric, also known as isocoric process, so the final temperature will be the water temperature at a saturated steam of v=0.05m3/kg, which is obtained by using steam tables for water, by linear interpolation. As follows, using table A-4 of the Cengel book 7th Edition:

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T=\frac{T2-T1}{v2-v1} x(v-v1)+T1=\frac{250.35°C-242.56°C}{0.049779m3/kg-0.057061m3/kg}x(0.05m3/kg-0.057061m3/kg)+242.56°C=250.11°C

The process work is zero because there is no change in volume during heating:

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W=process work

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2 years ago
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