The energy (in eV) possessed by a neon atom at `27^@ C` is

Options

A. `1.72 xx 10^(-3)`
B. `4.75 xx 10^(-4)`
C. `3.88 xx 10^(-2)`
D. `3.27 xx 10^(-5)`

Answer

To find the energy possessed by a neon atom at \(27^\circ C\), we can follow these steps:

### Step 1: Convert the temperature from Celsius to Kelvin
The formula to convert Celsius to Kelvin is:
\[ T(K) = T(°C) + 273 \]
Given \(T = 27^\circ C\):
\[ T = 27 + 273 = 300 \, K \]

### Step 2: Use the formula for the energy of a gas particle
The energy \(E\) possessed by a gas particle is given by the formula:
\[ E = \frac{3}{2} k T \]
where \(k\) is the Boltzmann constant.

### Step 3: Substitute the values into the energy formula
The value of the Boltzmann constant \(k\) is:
\[ k = 1.38 \times 10^{-23} \, \text{J/K} \]
Now substituting \(k\) and \(T\) into the energy formula:
\[ E = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 300 \]

### Step 4: Calculate the energy in Joules
Calculating the above expression:
\[ E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 \]
\[ E = \frac{3}{2} \times 4.14 \times 10^{-21} \]
\[ E = 6.21 \times 10^{-21} \, \text{J} \]

### Step 5: Convert energy from Joules to electron volts
To convert Joules to electron volts, we use the conversion factor:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Now, converting the energy:
\[ E_{eV} = \frac{6.21 \times 10^{-21}}{1.6 \times 10^{-19}} \]
Calculating this gives:
\[ E_{eV} = 3.88 \times 10^{-2} \, \text{eV} \]

### Final Result
The energy possessed by a neon atom at \(27^\circ C\) is approximately:
\[ E \approx 0.0388 \, \text{eV} \]

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