The surface of sodium metal is irradiated with radiation of wavelength $$x$$ nm. The kinetic energy of ejected electrons is $$2.8 \times 10^{-20}$$ J. The work function of sodium is 2.3 eV. The value of $$x$$ is _____ $$\times 10^2$$ nm. (Nearest integer) (Given: $$h = 6.6 \times 10^{-34}$$ J s; $$1$$ eV $$= 1.6 \times 10^{-19}$$ J; $$c = 3.0 \times 10^8$$ m s$$^{-1}$$)

The photo-electric equation relates the photon energy to the work function and the maximum kinetic energy: $$E_{\text{photon}} = \phi + K_{\text{max}}$$.

Given data:
• Work function of Na, $$\phi = 2.3\ \text{eV}$$
• Kinetic energy of emitted electrons, $$K_{\text{max}} = 2.8 \times 10^{-20}\ \text{J}$$
• Planck constant, $$h = 6.6 \times 10^{-34}\ \text{J s}$$
• Speed of light, $$c = 3.0 \times 10^{8}\ \text{m s}^{-1}$$
• $$1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$$

First convert the work function to joules:
$$\phi = 2.3\ \text{eV} \times 1.6 \times 10^{-19}\ \frac{\text{J}}{\text{eV}} = 3.68 \times 10^{-19}\ \text{J}$$

Total photon energy required:
$$E_{\text{photon}} = \phi + K_{\text{max}} = 3.68 \times 10^{-19} + 2.8 \times 10^{-20}$$ $$= 3.68 \times 10^{-19} + 0.28 \times 10^{-19} = 3.96 \times 10^{-19}\ \text{J}$$

For a photon, $$E_{\text{photon}} = \dfrac{hc}{\lambda}$$, so
$$\lambda = \dfrac{hc}{E_{\text{photon}}}$$ $$= \dfrac{(6.6 \times 10^{-34})(3.0 \times 10^{8})}{3.96 \times 10^{-19}}$$ $$= \dfrac{19.8 \times 10^{-26}}{3.96 \times 10^{-19}}$$ $$= 5.0 \times 10^{-7}\ \text{m}$$

Convert metres to nanometres: $$1\ \text{m} = 10^{9}\ \text{nm}$$, hence
$$\lambda = 5.0 \times 10^{-7}\ \text{m} = 5.0 \times 10^{-7} \times 10^{9}\ \text{nm} = 5.0 \times 10^{2}\ \text{nm}$$

The wavelength is therefore $$\mathbf{5 \times 10^2\ \text{nm}}$$. The question asks for the integer multiplying $$10^2\ \text{nm}$$, so the required value is

5.