4. The average thermal energy of a helium
atom at the temperature of the sun (6000K)
2) 3.67x10-19)
1) 422x10-195
3) 1.24x10-19
4) 2.53x10-19J
Answers
Answer:
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Explanation:
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(i) At room temperature, T=27
o
C=300K
Average thermal energy =(3/2)kT
Where, k is Boltzmann constant =1.38×10
−23
m
2
kgs
−2
K
−1
∴(3/2)kT=(3/2)×1.38×10
−38
×300
=6.21×10
−21
J
Hence, the average thermal energy of a helium atom at room temperature of 27
o
C is 6.21×10
−21
J.
(ii) On the surface of the sun, T=6000K
Average thermal energy =(3/2)kT
=(3/2)×1.38×10
−38
×6000
=1.241×10
−19
J
Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241×10
−19
J.
(iii) At temperature, T=10
7
K
Average thermal energy =(3/2)kT
=(3/2)×1.38×10
−23
×10
7
=2.07×10
−16
J
Hence, the average thermal energy of a helium atom at the core of a star is 2.07×10
−16
J.
Answer:
The Average Thermal energy of a helium atom at a temperature of the sun is 1.24 x 10^-19 .