#### Answer

The longest wavelength of radiation that possesses the necessary energy to break the bond is $494.3nm$.
A radiation with that wavelength is a form of visible light.

#### Work Step by Step

*Strategy:
Since we already know the minimum energy in $kJ/mol$ required to break the bond ($E_{min}$), we can find the minimum energy required to break a single bond ($E_{b, min}$). From there, we could deduce the longest wavelength of radiation that possesses the necessary energy to break the bond ($\lambda_{max}$).
- Step 1: Find the minimum energy required to break a single bond $E_{b, min}$
We have $E_{min}=242kJ/mol=2.42\times10^5J/mol$
and the number of bonds considered here: $N=1 mol Cl_2\approx6.022\times10^{23}$ bonds.
$E_{b, min}=\frac{E_{min}}{N}=\frac{2.42\times10^5}{6.022\times10^{23}}\approx4.019\times10^{-19}J/bond$
- Step 2: Find the longest wavelength of radiation having the necessary energy to break the bond $\lambda_{max}$
We would consider this formula $$E_{b}=h\times\nu=\frac{h\times c}{\lambda}$$
$E_{b}$ : the energy required to break a single bond
$h$ : Planck's constant ($h\approx6.626\times10^{-34}J.s$)
$\nu$ : frequency of radiation
$c$ : speed of light in a vacuum ($c\approx2.998\times10^8m/s$)
$\lambda$ : wavelength of radiation
Because both $h$ and $c$ are constants, and $E_{b}$ and $\lambda$ are inversely proportional, the minimum value of $E_b$ would require the maximum value of $\lambda$ (to break the bond). Therefore: $$E_{b, min}=\frac{h\times c}{\lambda_{max}}$$
In other words,
$\lambda_{max}=\frac{h\times c}{E_{b, min}}= \frac{(6.626\times10^{-34})\times(2.998\times10^8)}{4.019\times10^{-19}}\approx4.943\times10^{-7}m\approx494.3nm$
A radiation with a wavelength of $494.3nm$ lies in the wavelength range of visible light. Therefore, this radiation is a form of visible light.