JEE Chemistry 2024 Question with Solution

Given: Wavelength of electromagnetic radiation is $242 \, \text{nm}$.

Find: Ionization energy of sodium in $\text{kJ/mol}$.

The energy of one photon is given by

$$E = \frac{1240}{\lambda(\text{nm})} \, \text{eV}$$

Substituting $\lambda = 242 \, \text{nm}$,

$$E = \frac{1240}{242} \, \text{eV}$$ $$E = 5.12 \, \text{eV}$$

Convert this energy to joules per atom:

$$E = 5.12 \times 1.6 \times 10^{-19} \, \text{J/atom}$$ $$E = 8.198 \times 10^{-19} \, \text{J/atom}$$

Now convert joules per atom to $\text{kJ/mol}$ using Avogadro's number:

$$8.198 \times 10^{-19} \times 6.022 \times 10^{23} = 494 \, \text{kJ/mol}$$

Therefore, the ionization energy is $494 \, \text{kJ/mol}$. The correct option is C.

Given: Wavelength is $242 \, \text{nm}$.

Find: Ionization energy in $\text{kJ/mol}$.

Use the photon-energy relation first:

$$E = \frac{1240}{242} = 5.12 \, \text{eV}$$

Then convert $\text{eV}$ to molar energy using

$$1 \, \text{eV per particle} \approx 96.5 \, \text{kJ/mol}$$

So,

$$E = 5.12 \times 96.5 \, \text{kJ/mol}$$ $$E \approx 494 \, \text{kJ/mol}$$

This works because $\text{eV}$ is an energy per particle unit, and multiplying by the conversion factor directly gives energy per mole.

Therefore, the correct option is C.