Find the temperature at which the everage thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state `n = 3` state hydrogen can now emit rod light of wavelength `653.1 nm`because of maxwellan distribution of speeds a hydrogen sample emits red light at temperature much lower than that obtained from this problem Asuume that hydrogen that hydrogen molecules dissociate into atoms

Explanation

To find the temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state \( n = 3 \) to the ground state \( n = 1 \), we can follow these steps:

### Step 1: Calculate the Energy Required to Excite the Hydrogen Atom
The energy required to take a hydrogen atom from the \( n = 3 \) state to the ground state \( n = 1 \) can be calculated using the formula:

\[ E = 13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( n_1 = 1 \) (ground state)
- \( n_2 = 3 \)

Substituting the values:

\[ E = 13.6 \, \text{eV} \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 13.6 \, \text{eV} \left( 1 - \frac{1}{9} \right) = 13.6 \, \text{eV} \left( \frac{8}{9} \right) \]

Calculating this gives:

\[ E = 13.6 \times \frac{8}{9} \approx 12.09 \, \text{eV} \]

### Step 2: Relate Average Thermal Kinetic Energy to Temperature
The average thermal kinetic energy of a gas is given by:

\[ KE = \frac{3}{2} k T \]

Where:
- \( k \) is the Boltzmann constant \( (8.62 \times 10^{-5} \, \text{eV/K}) \)
- \( T \) is the temperature in Kelvin.

### Step 3: Set the Kinetic Energy Equal to the Excitation Energy
To find the temperature \( T \) at which the average thermal kinetic energy equals the energy required to excite the hydrogen atom, we set:

\[ \frac{3}{2} k T = E \]

Substituting \( E \):

\[ \frac{3}{2} k T = 12.09 \, \text{eV} \]

### Step 4: Solve for Temperature \( T \)
Rearranging the equation to solve for \( T \):

\[ T = \frac{2 \times E}{3k} \]

Substituting the values:

\[ T = \frac{2 \times 12.09 \, \text{eV}}{3 \times 8.62 \times 10^{-5} \, \text{eV/K}} \]

Calculating this gives:

\[ T = \frac{24.18}{2.586 \times 10^{-4}} \approx 9.35 \times 10^4 \, \text{K} \]

### Final Answer
Thus, the temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state \( n = 3 \) is approximately:

\[ T \approx 9.35 \times 10^4 \, \text{K} \]