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Found in: Page 669

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Integrated Concepts The temperature near the centre of the Sun is thought to be $$15{\rm{ }}million$$ degrees Celsius $$\left( {1.5 \times {{10}^7}^o{\rm{ }}C} \right)$$. Through what voltage must a singly charged ion be accelerated to have the same energy as the average kinetic energy of ions at this temperature?

Through $$V = 1.94 \times {10^3}\;V$$ a singly charged ion can be accelerated to have the same energy as the average kinetic energy of ions.

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Given Information

• Temperature near centre of the sun – $$\left( {1.5 \times {{10}^7}^o{\rm{ }}C} \right)$$

Concept Introduction

The Celsius and Kelvin temperature scales are related by the formula $$T = {T_C} + 273.15...(1)$$.

$${T_C}$$ is the temperature measurement in Celsius, while $$T$$ is the temperature reading in Kelvin.

The expression $$KE = \frac{3}{2}{k_B}T...(3)$$ relates the average translational kinetic energy per molecule of a gas, KE, to its temperature, $$T$$.

Where $${k_B} = 1.38 \times {10^{ - 23}}{m^2} \cdot kg/{s^2}$$, $$T$$ is expressed in kelvins and $${\rm{K}}$$ is the Boltzmann constant.

Electric Potential Energy: If a particle with charge $$q$$ is positioned at a location where a charged object's electric potential is $$V$$, the particle-object system's electric potential energy is -

$$PE = qV...(3)$$

Conservation of Energy Principle: The conservation of energy principle states that the sum of a system's initial energies plus any external forces' work on the system equals the system's final energies.

$${E_i} + W = {E_f}...(4)$$

Converting the temperature

Convert the temperature near the centre of the Sun for the Celsius scale to the Kelvin using Equation $$(1)$$:

$$T = {T_C} + 273.15\\ = 1.5 \times {10^7}^\circ C + 273.15\\ = 1.500027315 \times {10^7}\;K\ end$$

Deducing Kinetic Energy

The average kinetic energy of the ions at temperature $$T$$ is found from Equation $$(2)$$:

$$KE = \frac{3}{2}{k_B}T$$

Substitute numerical values:

$$KE = \frac{3}{2}\left( {1.38 \times {{10}^{ - 23}}\;{m^2} \times kg/{s^2} \times K} \right)\left( {1.500027315 \times {{10}^7}\;K} \right)\\ = 3.105 \times {10^{ - 16}}\;J\ end$$

Identifying Potential and Kinetic Energy

Ions and an electric field in an evacuated tube make up the system. We opt for the ions' initial position to be the $$V$$ field's zero level. Because the ions' mass is so little relative to their charge and won't ever reach statistically significant levels, ignore the potential energy caused by gravity. Additionally, the flight is over extremely quickly.

The system has neither kinetic energy or electric potential energy in its initial condition since the electric potential is zero (the ions are at rest). The system must have both positive kinetic energy and negative electric potential energy in its final state. Since the charge of the ions is negative and the electric potential in the final configuration is positive, we can calculate the electric potential energy $$P{E_f} = - eV$$ using Equation $$(3)$$. The electron's kinetic energy is written as $$K{E_f}$$.

Applying Energy Principle

The conservation of energy principle from Equation $$(4)$$ tells us, that since there are no nonconservative forces acting on the system, the initial energy of the system (which is zero) is equal to energy of the system in the final state. So,

$$0 = P{E_f} + K{E_f}\\0 = - eV + K{E_f}\\K{E_f} = eV\ end$$

Where we substitute $$K{E_f}$$we found above for $$KE$$ in order to find accelerating voltage $$V$$:

$$eV = KE$$

Solving for $$V$$, it is obtained –

$$V = \frac{{KE}}{e}$$

Entering the values for $$KE$$ and $$e$$, gives –

$$V = \frac{{3.105 \times {{10}^{ - 16}}\;J}}{{1.60 \times {{10}^{ - 19}}C}}\\ = 1.94 \times {10^3}\;\ end$$

Therefore, the voltage is obtained as $$V = 1.94 \times {10^3}\;V$$.

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