Question

The energy (in eV) possessed by a neon atom at $77^\circ$C is (Boltzmann constant, $k_B = 1.38 \times 10^{-23}$ J K$^{-1}$)

• A. $1.32 \times 10^{-3}$
• B. $3.20 \times 10^{-4}$
• C. $4.52 \times 10^{-2}$
• D. $3.88 \times 10^{-2}$

Hint:

Remember to always convert the temperature to Kelvin when using the Boltzmann constant. The average kinetic energy of a monatomic gas atom has 3 degrees of freedom, hence the factor $\frac{3}{2} k_B T$. The conversion between Joules and electron volts is crucial in atomic physics problems.

Answer 1

The Correct Option is C

Step 1: Convert the temperature to Kelvin. $$T(K) = T(^\circ \text{C}) + 273.15 = 77 + 273.15 = 350.15 \text{ K}$$
Step 2: Calculate the average kinetic energy in Joules. The average kinetic energy of a monatomic gas atom is $E = \frac{3}{2} k_B T$. $$E = \frac{3}{2} (1.38 \times 10^{-23} \text{ J K}^{-1}) (350.15 \text{ K})$$ $$E = 1.5 \times 1.38 \times 350.15 \times 10^{-23} \text{ J}$$ $$E = 724.8105 \times 10^{-23} \text{ J} = 7.248105 \times 10^{-21} \text{ J}$$
Step 3: Convert the energy from Joules to electron volts (eV). The conversion factor is $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$. $$E(\text{eV}) = \frac{E(\text{J})}{1.602 \times 10^{-19} \text{ J/eV}} = \frac{7.248105 \times 10^{-21}}{1.602 \times 10^{-19}} \text{ eV}$$ $$E(\text{eV}) = 4.5244 \times 10^{-2} \text{ eV}$$ Rounding to three significant figures, the energy is $4.52 \times 10^{-2}$ eV. Final Answer: The final answer is $\boxed{4.52 \times 10^{-2}}$