Ncert Exercises & Solutions

13.7 Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in case of a star).

At room temperature, T = 27°C = 300 K

Average thermal energy

= \frac{3}{2} k_{B} T

Where

k_{B} =

1.38 ×

10^{- 23}

m^{2} kg s^{- 2} k^{- 1}

E_{av} = \frac{3}{2} k_{B} T = \frac{3}{2} x 1 \cdot 38 \times 10^{- 23} \times 300

= 6.21 ×

10^{- 21}

Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 ×

10^{- 21}

On the surface of the sun, T = 6000 K

Average thermal energy

= \frac{3}{2} k_{B} T

= \frac{3}{2} x 1 \cdot 38 \times 10^{- 23} \times 6000

= 1.241 ×

10^{- 19}

Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 ×

10^{- 19}

At temperature, T =

10^{7}

Average thermal energy

= \frac{3}{2} k_{B} T

= \frac{3}{2} x 1 \cdot 38 \times 10^{- 23} \times 10^{7}

= 2.07 × 1

0^{- 16}

Hence, the average thermal energy of a helium atom at the core of a star is 2.07 ×

10^{- 16}

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