Chapter 25: Q. 43 (page 711)

A proton’s speed as it passes point A is 50,000 m/s. It follows the trajectory shown in FIGURE P25.43. What is the proton’s speed at point B? A proton’s speed as it passes point A is 50,000 m/s. It follows the trajectory shown in FIGURE P25.43. What is the proton’s speed at point B?

Short Answer

Expert verified

The value of proton’s speed at the point B is $1.01 \times 10^{- 5} m / s$

Step by step solution

Speed of photon

The potential energy and kinetic energy at the A point

$K i n e c t i c e n e r g y i s K_{A} = \frac{1}{2} m {v^{2}}_{A} p o t e n t i a l e n e r g y i s U_{A} = e V_{A}$

The potential energy and kinetic energy at the B point

$K i n e c t i c e n e r g y i s K_{B} = \frac{1}{2} m {v^{2}}_{B} p o t e n t i a l e n e r g y i s U_{B} = e V_{B}$

$e V_{A} + \frac{1}{2} m {v^{2}}_{A} = e V_{B} + \frac{1}{2} m {v^{2}}_{B} v_{B} = \sqrt{{v^{2}}_{A} + \frac{2 \times e (V_{A} - V_{B})}{m}} v_{A} = 5000 m / s e = 1.6 \times 10^{- 19} C V_{A} = 30 V V_{B} = - 10 V v_{B} = 1.01 \times 10^{5} m / s$

Calculation

On the basis of conservation theory

$K_{A} + U_{A} = K_{B} + U_{B}$

Applying all data

role="math" localid="1648296227023" $\begin{array}{rcl} V_{B} & = & \sqrt{{V_{A}}^{2} + \frac{2 e (V_{A} - V_{B})}{m}} \\ = & \sqrt{{(50000 m / s)}^{2} + \frac{2 (1.6 \times 10^{- 19} C) (30 V - (- 10 V))}{(1.67 \times 10^{- 27} k g)}} \\ = & 1.008 \times 10^{5} m / s \end{array}$

Therefore, the speed of the proton is $1.01 \times 10^{5} m / s$

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A proton’s speed as it passes point A is 50,000 m/s. It follows the trajectory shown in FIGURE P25.43. What is the proton’s speed at point B? A proton’s speed as it passes point A is 50,000 m/s. It follows the trajectory shown in FIGURE P25.43. What is the proton’s speed at point B?

Short Answer

Step by step solution

Speed of photon

Calculation

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