NEET Physics (general) MCQ Question

Using the relation $\lambda = \frac{hc}{\phi}$, where $\phi = 3.0 \text{ eV} = 3.0 \times 1.6 \times 10^{-19} \text{ J}$, $h = 6.63 \times 10^{-34} \text{ J s}$, and $c = 3 \times 10^8 \text{ m/s}$, we calculate $\lambda = \frac{(6.63 \times 10^{-34} \text{ J s}) (3 \times 10^8 \text{ m/s})}{3.0 \times 1.6 \times 10^{-19} \text{ J}} = 414 \text{ nm}$.