Question

The first ionization energy of chlorine is 1.25x103 kJ/mol. What is the wavelength of light in units of meters (absorbed) which can ionize a chlorine atom? A. 9.57x10–8 B. 3.13x1015 C. 1.89x1036 D. 3.13x1012 E. 9.57x10–5

Answer 1

Answer:

The correct option is: A. 9.57 × 10⁻⁸ m

Explanation:

Given:  First molar ionization energy of chlorine (Cl) = 1.25 × 10³ kJ/mol

Since, 1 kJ = 10³ J

∴ First molar ionization energy of chlorine (Cl) = 1.25 × 10³ kJ/mol = 1.25 × 10³ × 10³ J/mol = 1.25 × 10⁶ J/mol

Now, for a single Cl atom,

the first ionization energy of a Cl atom: $E (J) = \frac{First\: molar\: ionization\: energy (J/mol)}{Avogadro\: constant (N_{A})}$

$As\: we\: know,\: Avogadro\: constant: \: N_{A} = 6.022 \times 10^{23} mol^{-1}$

$\therefore E = \frac{1.25\times 10^{6} J/mol}{6.022 \times 10^{23} mol^{-1}} = 2.076\times 10^{-18} J$

Now, to find the wavelength of light absorbed to ionize the Cl atom, we use the Planck equation:

$E = h\nu =\frac{hc}{\lambda}$

$Here, c:\: speed\: of\: light = 3 \times 10^{8}\: m/s$

$h:\: Planck\: constant = 6.626\times 10^{-34}\: J.s$

$\nu: frequency\: of\: light$

$\lambda:\: wavelength\: of\: light = ?$

Therefore, the wavelength of light (λ) that can ionize the Cl atom: $\lambda = \frac{hc}{E}$

$\therefore \lambda = \frac{(6.626\times 10^{-34}\: J.s)\times 3 \times 10^{8}\: m/s}{2.076\times 10^{-18} J} = 9.575 \times 10^{-8}\: m$

 

Therefore, the wavelength of light (λ) required to ionize a chlorine atom = 9.57 × 10⁻⁸ m