A He atom at 300 K is released from the surface of the earth to travel upwards. Assuming that it undergoes no collision with other molecules, how high will it be before coming to the rest?

Options

A. 9.53 m
B. 9.5 m
C. 953 m
D. `9.53xx10^(4)` m

Explanation

To solve the problem of how high a helium atom will rise before coming to rest when released from the surface of the Earth, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the Energy Conservation Principle
The total mechanical energy of the helium atom is conserved. Initially, it has kinetic energy (KE) when released, and as it rises, this kinetic energy is converted into gravitational potential energy (PE) until it comes to rest at its maximum height.

### Step 2: Write the Equations for Kinetic and Potential Energy
The kinetic energy (KE) of the helium atom can be expressed as:
\[ KE = \frac{3}{2} k_B T \]
Where:
- \( k_B \) is the Boltzmann constant (\( 1.38 \times 10^{-23} \, \text{J/K} \))
- \( T \) is the temperature in Kelvin (300 K)

The potential energy (PE) at height \( h \) is given by:
\[ PE = mgh \]
Where:
- \( m \) is the mass of the helium atom
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))
- \( h \) is the height

### Step 3: Calculate the Mass of the Helium Atom
The mass of a helium atom is approximately \( 4 \times 1.67 \times 10^{-27} \, \text{kg} \) (since helium has 4 nucleons: 2 protons and 2 neutrons).
\[ m = 4 \times 1.67 \times 10^{-27} = 6.68 \times 10^{-27} \, \text{kg} \]

### Step 4: Set Kinetic Energy Equal to Potential Energy
At the maximum height, all kinetic energy will have been converted to potential energy:
\[ \frac{3}{2} k_B T = mgh \]

### Step 5: Substitute Known Values
Substituting the known values into the equation:
\[ \frac{3}{2} (1.38 \times 10^{-23}) (300) = (6.68 \times 10^{-27}) (10) h \]

### Step 6: Solve for Height \( h \)
First, calculate the left side:
\[ \frac{3}{2} (1.38 \times 10^{-23}) (300) = 6.21 \times 10^{-21} \, \text{J} \]

Now, substitute this into the equation:
\[ 6.21 \times 10^{-21} = (6.68 \times 10^{-27}) (10) h \]
\[ 6.21 \times 10^{-21} = 6.68 \times 10^{-26} h \]

Now, solve for \( h \):
\[ h = \frac{6.21 \times 10^{-21}}{6.68 \times 10^{-26}} \]
\[ h \approx 9.31 \times 10^{4} \, \text{m} \]

### Final Answer
The height \( h \) that the helium atom will reach before coming to rest is approximately:
\[ h \approx 9.31 \times 10^{4} \, \text{meters} \]

---