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  • Find out the temperature at which the average kinetic energy of a gas molecule will be equal to the energy gained by an electron on acceleration across a potential difference of 1v. Given Boltzman

Find out the temperature at which the average kinetic energy of a gas molecule will be equal to the energy gained by an electron on acceleration across a potential difference of 1v. Given Boltzman constant=1.38 \times 10^{-23} \mathrm{~J} \cdot \mathrm{K}^{-1}; change of an election =1.6 \times 10^{-19} \mathrm{c}

Option: 1

7456^{\circ} \mathrm{C}


Option: 2

7457


Option: 3

7729


Option: 4

7656^{\circ} \mathrm{C}


Answers (1)

best_answer

Energy gained by electron

\begin{aligned} & =1 e^{V}=\left(1.6 \times 10^{-19} \mathrm{C}\right) \times 1 \mathrm{~V}=1.6 \times 10^{-19} \mathrm{~J} \end{aligned}

Average kinetic energy of a gas molecule

\begin{aligned} = & \frac{3}{2} \mathrm{KT}=\frac{3}{2} \times\left(1.38 \times 10^{-23}\right) \times T \mathrm{~J} \\ \therefore & 1.6 \times 10^{-19}=\frac{3}{2} \times\left(1.38 \times 10^{-23}\right) \times T \\ \text { or } T & =\frac{2}{3} \times \frac{1.6 \times 10^{-19}}{1.38 \times 10^{-23}}=7729 \mathrm{~K}=7456^{\circ} \mathrm{C} \end{aligned}

 

Posted by

Divya Prakash Singh

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