The work function of sodium metal is $$4.41 \times 10^{-19}$$ J. If photons of wavelength 300 nm are incident on the metal, the kinetics energy of the ejected electrons will be $$(h = 6.63 \times 10^{-34}$$ J s; $$c = 3 \times 10^8$$ m s$$^{-1})$$ ___________ $$\times 10^{-21}$$ J.

For photoelectric emission we use Einstein’s photoelectric equation, which states

$$E_{\text{photon}} = \phi \;+\; K.E.$$

where $$E_{\text{photon}}$$ is the energy of the incident photon, $$\phi$$ is the work-function of the metal, and $$K.E.$$ is the kinetic energy of the emitted electron.

First we calculate the photon energy from its wavelength. The energy-wavelength relation is

$$E_{\text{photon}} \;=\; \dfrac{h\,c}{\lambda}.$$

We have the numerical values

$$h = 6.63 \times 10^{-34}\,\text{J s}, \qquad c = 3.00 \times 10^{8}\,\text{m s}^{-1}, \qquad \lambda = 300\,\text{nm} = 300 \times 10^{-9}\,\text{m}.$$

Substituting these in the formula gives

$$E_{\text{photon}} = \dfrac{(6.63 \times 10^{-34})\,(3.00 \times 10^{8})}{300 \times 10^{-9}}\;\text{J}.$$

Multiplying the numerators,

$$6.63 \times 3.00 = 19.89, \quad\text{and}\quad 10^{-34}\times 10^{8}=10^{-26},$$

so the numerator becomes

$$19.89 \times 10^{-26}\;\text{J m}.$$

For the denominator, write

$$300 \times 10^{-9} = (3.00 \times 10^{2}) \times 10^{-9} = 3.00 \times 10^{-7}\;\text{m}.$$

Now divide:

$$E_{\text{photon}} = \dfrac{19.89 \times 10^{-26}}{3.00 \times 10^{-7}}\;\text{J} = \left(\dfrac{19.89}{3.00}\right) \times 10^{-26+7}\;\text{J}.$$

Calculating the numeric and exponent parts separately,

$$\dfrac{19.89}{3.00} = 6.63, \qquad -26 + 7 = -19,$$

hence

$$E_{\text{photon}} = 6.63 \times 10^{-19}\;\text{J}.$$

The work-function of sodium is given as

$$\phi = 4.41 \times 10^{-19}\;\text{J}.$$

Using Einstein’s equation, the kinetic energy of the photoelectrons is

$$K.E. = E_{\text{photon}} - \phi = (6.63 \times 10^{-19}) - (4.41 \times 10^{-19})\;\text{J}.$$

Subtracting the coefficients while keeping the common power of ten,

$$6.63 - 4.41 = 2.22,$$

so

$$K.E. = 2.22 \times 10^{-19}\;\text{J}.$$

Finally, the problem asks for the answer in units of $$\times 10^{-21}\;\text{J}.$$ To rewrite the kinetic energy accordingly, note that

$$2.22 \times 10^{-19}\;\text{J} = 2.22 \times 10^{2} \times 10^{-21}\;\text{J} = 222 \times 10^{-21}\;\text{J}.$$

So, the answer is $$222$$.