If the work function of a metal is $$6.63 \times 10^{-19}$$ J, the maximum wavelength of the photon required to remove a photoelectron from the metal is ______ nm. Nearest integer
[Given : $$h = 6.63 \times 10^{-34}$$ J s, and $$c = 3 \times 10^{8}$$ m s$$^{-1}$$]

We need to find the maximum wavelength of a photon required to remove a photoelectron from a metal with work function $$\phi = 6.63 \times 10^{-19}$$ J. The photoelectric effect requires the photon energy to be at least equal to the work function of the metal; the minimum energy corresponds to the maximum wavelength since energy and wavelength are inversely related. At the threshold (minimum energy = work function), $$E_{photon} = \phi$$.

The energy of a photon is given by $$E = \frac{hc}{\lambda}$$, so at threshold $$\frac{hc}{\lambda_{max}} = \phi$$. Solving for $$\lambda_{max}$$ gives $$\lambda_{max} = \frac{hc}{\phi}$$.

Using $$h = 6.63 \times 10^{-34}$$ J s, $$c = 3 \times 10^8$$ m/s, and $$\phi = 6.63 \times 10^{-19}$$ J, we get $$\lambda_{max} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6.63 \times 10^{-19}}$$, which simplifies to $$\lambda_{max} = \frac{6.63 \times 3 \times 10^{-34+8}}{6.63 \times 10^{-19}}$$, then to $$\lambda_{max} = \frac{3 \times 10^{-26}}{10^{-19}}$$, and hence $$\lambda_{max} = 3 \times 10^{-7} \text{ m}$$.

Converting to nanometers: $$\lambda_{max} = 3 \times 10^{-7} \text{ m} = 300 \times 10^{-9} \text{ m} = 300 \text{ nm}$$. The correct answer is 300 nm.