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प्रश्न
Estimate the average thermal energy of a helium atom at room temperature (27 °C).
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उत्तर
At room temperature, T = 27°C = 300 K
Average thermal energy = `3/2` kT
Where k is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1
`therefore 3/2kT = 3/2 xx 1.38 xx 10^(-38) xx 300`
= 6.21 × 10–21J
Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 × 10–21 J.
संबंधित प्रश्न
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