Respuesta :
Answer: The question is incomplete as some other details are missing. Here is the complete question ; A helium atom falls in a vacuum . The mass of a helium atom is 6.64 x 10-27 kg . (a) From what height must it fall so that its translational kinetic energy at the bottom equals the average translational kinetic energy of a helium molecule at 300 K? (b) At what temperature would the average speed of helium atoms equal the escape speed from the Earth, 1.12 x 104 m/s.
a) Height = 95.336Km
b) Temperature = 20118.87K
Explanation:
The detailed steps and calculations is shown in the attached file.
The helium atom must fall from 95.336 km to achieve average translational kinetic energy at 300 K.
What is Translational kinetic energy?
It is the required energy to accelerate the particle or object from the rest to achieve the given velocity.
The height can be calculated by the formula,
[tex]h = \dfrac 23 \times \dfrac {kt}{mg}[/tex]
Where
h - height =?
k - constant =[tex]1.38\times 10^{-23}[/tex] J/k
t - temperature = 300 k
m - mass
g - gravitational acceleration = 9.8 m/s^2
Put the values in the formula,
[tex]h = \dfrac 23 \times \dfrac {1.38\times 10^{-23}\times 300 }{6.64 \times 10^{-27}\times 9.8}\\\\h = 95335.4 {\rm\ m \ \ or}\\\\h = 95.336\rm \ km[/tex]
Therefore, the helium atom must fall from 95.336 km to achieve average translational kinetic energy at 300 K.
Learn more about translational kinetic energy:
https://brainly.com/question/25612167
