The first excited state of hydrogen atom is `10.2 eV` above its ground state. The temperature is needed to excite hydrogen atoms to first excited level is

Options

A. `7.9 xx 10^(4)K`
B. `3.5 xx 10^(4)K`
C. `5.8 xx 10^(4) K`
D. `14 xx 10^(4)K`

Explanation

To find the temperature needed to excite hydrogen atoms to the first excited state, we can follow these steps:

### Step 1: Understand the energy difference
The energy difference between the ground state and the first excited state of the hydrogen atom is given as \(10.2 \, \text{eV}\).

### Step 2: Convert energy from eV to Joules
To perform calculations in SI units, we need to convert the energy from electron volts (eV) to joules (J). The conversion factor is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus, the energy in joules is:
\[ E = 10.2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.632 \times 10^{-18} \, \text{J} \]

### Step 3: Use the formula for kinetic energy
The average kinetic energy of a particle in a gas is given by:
\[ KE = \frac{3}{2} k T \]
where \(k\) is the Boltzmann constant (\(1.38 \times 10^{-23} \, \text{J/K}\)) and \(T\) is the temperature in Kelvin.

### Step 4: Set the kinetic energy equal to the energy required for excitation
We set the kinetic energy equal to the energy required to excite the hydrogen atom:
\[ \frac{3}{2} k T = E \]
Substituting \(E\) from Step 2:
\[ \frac{3}{2} (1.38 \times 10^{-23}) T = 1.632 \times 10^{-18} \]

### Step 5: Solve for temperature \(T\)
Rearranging the equation to solve for \(T\):
\[ T = \frac{2 \times 1.632 \times 10^{-18}}{3 \times 1.38 \times 10^{-23}} \]
Calculating the numerator:
\[ 2 \times 1.632 \times 10^{-18} = 3.264 \times 10^{-18} \]
Calculating the denominator:
\[ 3 \times 1.38 \times 10^{-23} = 4.14 \times 10^{-23} \]
Now substituting back into the equation for \(T\):
\[ T = \frac{3.264 \times 10^{-18}}{4.14 \times 10^{-23}} \approx 7.9 \times 10^{4} \, \text{K} \]

### Final Answer
The temperature needed to excite hydrogen atoms to the first excited level is approximately \(7.9 \times 10^{4} \, \text{K}\).

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