Calculate the temperature (in K) at which kinetic energy of mono-atomic gaseous molecule is equal to 0.414 eV

Correct Option: 3199 K

Step-by-Step Explanation:

To find the temperature at which the kinetic energy of a mono-atomic gaseous molecule is equal to 0.414 eV, we can use the kinetic theory of gases. The average kinetic energy of a single mono-atomic gas molecule at a absolute temperature T is given by the formula:

$${\text{KE}}_{\text{molecule}}=\frac{3}{2}{k}_{\text{B}}T=\frac{3}{2}\left(\frac{R}{N_{\text{A}}}\right)T$$
where:

• $k_{\text{B}}$ is the Boltzmann constant.
• $R$ is the universal gas constant ($8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$).
• $N_{\text{A}}$ is Avogadro's number ($6.022Õ{10}^{23}{\text{mol}}^{-1}$).
• $T$ is the absolute temperature in Kelvin (K).

From the visual analysis of the provided image, we can see these relations set up directly as:

$${\text{(KE)}}_{\text{atom}}=\frac{3}{2}\left(\frac{\text{R}}{{\text{N}}_{\text{A}}}\right)\text{(T)}$$

1. Convert the Kinetic Energy from electron-volts (eV) to Joules (J):
The given kinetic energy is $0.414\text{eV}$.
Since $1\text{eV}=1.6Õ{10}^{-19}\text{J}$:

$$\text{KE}=0.414Õ1.6Õ{10}^{-19}\text{J}$$

2. Substitute the values into the equation:
Substituting the energy and constant values into the formula shown in the image:

$$0.414Õ1.6Õ{10}^{-19}=\frac{3}{2}Õ\left(\frac{8.314}{6.022Õ{10}^{23}}\right)ÕT$$

3. Solve for Temperature (T):
Rearranging the equation to solve for $T$:

$$T=\frac{\left(0.414\right)Õ1.6Õ2Õ6.022Õ{10}^4}{3Õ8.314}$$
Notice that ${10}^{-19}Õ{10}^{23}={10}^4$, which is reflected in the numerator as shown in the derivation steps in the image.

Calculating the numerator:
$0.414Õ1.6Õ2Õ6.022Õ10000=79779.456$

Calculating the denominator:
$3Õ8.314=24.942$

Dividing the two results:
$$T\approx 3198.59\text{K}$$
Rounding to the nearest whole number yields:

$$T\approx 3199\text{K}$$

Therefore, the temperature at which the kinetic energy of a mono-atomic gas molecule equals 0.414 eV is approximately 3199 K.