What is the wavelength in nanometer of radiation that has energy content 1×10³KJ/mol?

First, convert the given energy from kilojoules per mole to joules per mole.

$$E_{mol} = 1 \times 10^3 \text{ kJ/mol} \times \frac{1000 \text{ J}}{1 \text{ kJ}}$$ $$E_{mol} = 1 \times 10^6 \text{ J/mol}$$

To find the energy of a single photon ($E_{photon}$), divide the molar energy by Avogadro's number.

$$E_{photon} = \frac{E_{mol}}{N_A}$$ $$E_{photon} = \frac{1 \times 10^6 \text{ J/mol}}{6.022 \times 10^{23} \text{ mol}^{-1}}$$ $$E_{photon} = 1.66057788 \times 10^{-18} \text{ J}$$

The energy of a photon is related to its wavelength ($\lambda$) by the equation $E_{photon} = \frac{hc}{\lambda}$. Rearranging this equation to solve for wavelength:

$$\lambda = \frac{hc}{E_{photon}}$$ $$\lambda = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.00 \times 10^8 \text{ m/s})}{1.66057788 \times 10^{-18} \text{ J}}$$ $$\lambda = \frac{1.9878 \times 10^{-25} \text{ J m}}{1.66057788 \times 10^{-18} \text{ J}}$$ $$\lambda = 1.19704 \times 10^{-7} \text{ m}$$

Since $1 \text{ m} = 10^9 \text{ nm}$, convert the wavelength from meters to nanometers.

$$\lambda = 1.19704 \times 10^{-7} \text{ m} \times \frac{10^9 \text{ nm}}{1 \text{ m}}$$ $$\lambda = 119.704 \text{ nm}$$

Rounding to one significant figure, as indicated by the given energy content ($1 \times 10^3 \text{ kJ/mol}$):

$$\lambda \approx 1 \times 10^2 \text{ nm}$$

The wavelength of the radiation is:

$$1 \times 10^2 \text{ nm}$$