Step 1: Use the direct eV-nanometre shortcut instead of SI units.
Planck's relation gives $E = \frac{hc}{\lambda}$, but converting $h$ and $c$ into SI units and then dividing eV out is slow. A quicker route many chemists use is to combine $hc$ and the eV conversion factor once and for all into a single constant:
\[ hc \approx 1240 \, \text{eV} \cdot \text{nm} \]
This number already has the joule to eV conversion folded in, so once $\lambda$ is written in nanometres the energy comes out directly in eV.
Step 2: Write the working formula.
With this shortcut the photon energy formula becomes:
\[ E(\text{eV}) = \frac{1240}{\lambda(\text{nm})} \]
Step 3: Substitute the given wavelength.
The wavelength given is $\lambda = 700$ nm, so:
\[ E = \frac{1240}{700} \]
Step 4: Carry out the division.
\[ E = 1.7714 \approx 1.77 \, \text{eV} \]
Step 5: Check the other options.
Option (2), 17.7 eV, is what you get by misplacing a decimal point in the division. Options (3) and (4), 4.3 eV and 14.3 eV, do not follow from this formula at all and correspond to using the wrong power of ten while converting joules to eV. None of them match the direct calculation.
Final Answer:
The energy of the 700 nm radiation is:
\[ \boxed{1.77 \, \text{eV}} \]