Electromagnetic radiation of wavelength 663 nm is just sufficient to ionise the atom of metal A. The ionization energy of metal A in `kJ "mol"^(-1)` is _______. (Rounded-off to the nearest integer) `[h = 6.63 xx 10^(-34) Js, c = 3.00 xx 10^8 ms^(-1) , N_A = 6.02 xx 10^23 "mol"^(-1)`]

Answer

To find the ionization energy of metal A in kJ/mol, we can use the formula that relates energy (E) to wavelength (λ):

\[ E = \frac{hc}{\lambda} \]

Where:
- \( h \) is Planck's constant \( (6.63 \times 10^{-34} \text{ Js}) \)
- \( c \) is the speed of light \( (3.00 \times 10^8 \text{ ms}^{-1}) \)
- \( \lambda \) is the wavelength in meters.

### Step 1: Convert the wavelength from nanometers to meters
Given that the wavelength \( \lambda = 663 \text{ nm} \):

\[ \lambda = 663 \text{ nm} = 663 \times 10^{-9} \text{ m} \]

### Step 2: Substitute the values into the energy formula
Now, substitute \( h \), \( c \), and \( \lambda \) into the energy formula:

\[ E = \frac{(6.63 \times 10^{-34} \text{ Js}) \times (3.00 \times 10^8 \text{ ms}^{-1})}{663 \times 10^{-9} \text{ m}} \]

### Step 3: Calculate the energy in joules
Calculating the numerator:

\[ E = \frac{(6.63 \times 3.00) \times 10^{-34} \times 10^8}{663 \times 10^{-9}} \]

Calculating \( 6.63 \times 3.00 = 19.89 \):

\[ E = \frac{19.89 \times 10^{-26}}{663 \times 10^{-9}} \]

Now, calculating \( 663 \times 10^{-9} = 6.63 \times 10^{-7} \):

\[ E = \frac{19.89 \times 10^{-26}}{6.63 \times 10^{-7}} \]

Now, performing the division:

\[ E \approx 2.997 \times 10^{-19} \text{ J} \]

### Step 4: Convert energy from joules to kilojoules
To convert joules to kilojoules, we divide by 1000:

\[ E \approx \frac{2.997 \times 10^{-19}}{1000} \approx 2.997 \times 10^{-22} \text{ kJ} \]

### Step 5: Convert energy per atom to energy per mole
To find the energy per mole, we multiply by Avogadro's number \( (N_A = 6.02 \times 10^{23} \text{ mol}^{-1}) \):

\[ E_{\text{per mole}} = 2.997 \times 10^{-19} \text{ J} \times 6.02 \times 10^{23} \text{ mol}^{-1} \]

Calculating this gives:

\[ E_{\text{per mole}} \approx 180.33 \text{ kJ/mol} \]

### Final Answer
Thus, the ionization energy of metal A is approximately:

**180 kJ/mol (rounded to the nearest integer)**

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