(a) Explanation Section 1: To determine: The root mean square speed of the deuteron. Answer: The root mean square speed of the deuteron is 1.93 × 10 6 m/s . Explanation: Given info: The nucleus of deuteron 1 2 H , the highest temperature is 300 million K and the speed of the light is 3 × 10 8 m/s . The expression of root mean square speed of the gas is, V r m s = 3 k T m (I) k is the Boltzmann constant T is the temperature of the gas m is the mass of the gas V r m s is the root mean square speed of the gas The mass of one deuteron due to presence of one proton and one neutron is 2 × 1836 × 9.1 × 10 − 31 kg and the value of Boltzmann constant is 1.38 × 10 − 23 . Substitute 300 million K for T , 1.38 × 10 − 23 for k and 2 × 1836 × 9.1 × 10 − 31 kg for m to find V r m s V r m s = 3 ( 1.38 × 10 − 23 ) ( 300 × 10 6 K ) 2 × 1836 × 9 (b) To determine The temperature of the plasma.

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ISBN: 9780133969290

Author: Hugh D. Young, Roger A. Freedman

Publisher: PEARSON

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Chapter 18, Problem 18.31E

(a)

To determine

The root mean square speed of the deuteron and the value is significant or not as a fraction of speed of light.

(b)

To determine

The temperature of the plasma.

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Chapter 18, Problem 18.30E

Chapter 18, Problem 18.32E

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The root mean square speed of the deuteron and the value is significant or not as a fraction of speed of light .

Solution Summary: The author explains the root mean square speed of the deuteron and the value is signification as a tion of light by 0.64%

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The root mean square speed of the deuteron and the value is significant or not as a fraction of speed of light.

The temperature of the plasma.

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