Define wave-particle duality. 2. Calculate the kinetic energy of an electron ejected from a metal surface if the incident light has a frequency of 6x1014 Hz and the work function of the metal is 6eV.

The maximum kinetic energy ($KE_{max}$) of an electron ejected during the photoelectric effect is given by Einstein's photoelectric equation:

$$KE_{max} = hf - \phi$$

where $h$ is Planck's constant, $f$ is the frequency of the incident light, and $\phi$ is the work function of the metal.

Given $\phi = 6 \text{ eV}$ and $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$.

$$\phi = 6 \text{ eV} \times \left(1.6 \times 10^{-19} \frac{\text{J}}{\text{eV}}\right)$$ $$\phi = 9.6 \times 10^{-19} \text{ J}$$

Given $h = 6.6 \times 10^{-34} \text{ J s}$ and $f = 6 \times 10^{14} \text{ Hz}$.

$$E = hf$$ $$E = (6.6 \times 10^{-34} \text{ J s}) \times (6 \times 10^{14} \text{ Hz})$$ $$E = 39.6 \times 10^{-20} \text{ J}$$ $$E = 3.96 \times 10^{-19} \text{ J}$$

Now, substitute the values of $E$ and $\phi$ into the photoelectric equation:

$$KE_{max} = E - \phi$$ $$KE_{max} = (3.96 \times 10^{-19} \text{ J}) - (9.6 \times 10^{-19} \text{ J})$$ $$KE_{max} = (3.96 - 9.6) \times 10^{-19} \text{ J}$$ $$KE_{max} = -5.64 \times 10^{-19} \text{ J}$$

Since the kinetic energy cannot be negative, this result indicates that the energy of the incident photon ($E = 3.96 \times 10^{-19} \text{ J}$) is less than the work function of the metal ($\phi = 9.6 \times 10^{-19} \text{ J}$). Therefore, no electrons will be ejected from the metal surface.

The kinetic energy of an ejected electron is $\boxed{\text{0 J}}$ because no electrons are ejected.