At which temperature will the translational kinetic energy of H-atom equal to that for H-atom of first line Lyman transition? (Given `N_(A)=6xx10^(23)`)

Options

A. (a) `780K`
B. (b) `1.32xx10^95`K
C. (c) `7.84xx10^(4)`K
D. (d) `1000K`

Answer

To find the temperature at which the translational kinetic energy of a hydrogen atom equals the energy associated with the first line of the Lyman transition, we can follow these steps:

### Step 1: Understand the Lyman Transition
The first line of the Lyman series corresponds to the transition of an electron in a hydrogen atom from the n=2 energy level to the n=1 energy level.

### Step 2: Calculate the Wavenumber (1/λ)
The wavenumber (1/λ) for the transition can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For hydrogen (Z=1), \(n_1 = 1\), and \(n_2 = 2\):
\[ \frac{1}{\lambda} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \]

### Step 3: Substitute the Rydberg Constant
The Rydberg constant \(R\) is approximately \(1.1 \times 10^7 \, \text{m}^{-1}\):
\[ \frac{1}{\lambda} = \frac{3}{4} \cdot 1.1 \times 10^7 = 8.25 \times 10^6 \, \text{m}^{-1} \]

### Step 4: Calculate the Energy of the Transition
The energy \(E\) associated with this transition can be calculated using:
\[ E = \frac{hc}{\lambda} \]
Where:
- \(h = 6.626 \times 10^{-34} \, \text{J s}\) (Planck's constant)
- \(c = 3 \times 10^8 \, \text{m/s}\)

Using \( \lambda = \frac{1}{\frac{1}{\lambda}} = \frac{1}{8.25 \times 10^6} \):
\[ E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{\frac{1}{8.25 \times 10^6}} = (6.626 \times 10^{-34})(3 \times 10^8)(8.25 \times 10^6) \]
Calculating this gives:
\[ E \approx 1.63 \times 10^{-18} \, \text{J} \]

### Step 5: Equate to Translational Kinetic Energy
The translational kinetic energy \(KE\) of a gas particle is given by:
\[ KE = \frac{3}{2} k_B T \]
Where \(k_B\) (Boltzmann constant) is:
\[ k_B = \frac{R}{N_A} \approx 1.38 \times 10^{-23} \, \text{J/K} \]

Setting the energies equal:
\[ \frac{3}{2} k_B T = 1.63 \times 10^{-18} \]

### Step 6: Solve for Temperature \(T\)
Rearranging gives:
\[ T = \frac{2 \cdot 1.63 \times 10^{-18}}{3 \cdot 1.38 \times 10^{-23}} \]
Calculating this yields:
\[ T \approx 7.84 \times 10^4 \, \text{K} \]

### Final Answer
The temperature at which the translational kinetic energy of a hydrogen atom equals that for the hydrogen atom of the first line Lyman transition is approximately:
\[ T \approx 7.84 \times 10^4 \, \text{K} \]

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