$E = hf$

where $f$ is the frequency and $h$ is the Planck’s constant.

The value of Planck’s constant is $h = 6.626 times 10^{-34}:J.s$.

The frequency is given as $f=1 times 10^{15}:Hz$. So the energy of the photon is given by

$E = hf = 6.626 times 10^{-34} times 1 times 10^{15} = 6.626 times 10^{-19}:J$

Its momentum is given by

$p = dfrac{h}{lambda}$

where $h$ is Planck’s constant and its value is given by $6.626 times 10^{-34}:J.s$

$p = dfrac{6.626 times 10^{-34}}{200 times 10^{-9}} = dfrac{6.626 times 10^{-34}}{2 times 10^{-7}}$

$p = dfrac{6.626}{2} times 10^{-34} times 10^7 = 3.313 times 10^{-27}: J.s .m^{-1} = 3.313 times 10^{-27}:Kg.m.s^{-1}$

$[ J = Kg. m^2. s^{-2} ]$

$W_0 = h f_0$

where $f_0$ is the cutoff frequency and $h (=6.626 times 10^{-34}:J.s)$ is Planck’s constant.

The value of the cutoff frequency of the metal is given as $f_0 = 1.5 times 10^{15}:Hz$.

$W_0 = 6.626 times 10^{-34} times 1.5 times 10^{15} = 6.626 times 1.5 times 10^{-34} times 10^{15} = 9.939 times 10^{-19}:J$

$K_{max} = hf – W_0$

where $h$ is the Planck’s constant, $f$ is the frequency and $W_0$ is the work function.

When $K_{max}$ is plotted as a function of $f$, the curve is the straight line with $h$ being the slope and $-W_0$ being the $y$-intercept value.

So the slope of the graph is Planck’s constant ($h$).

$K_{max} = hf – W_0$

where $h$ is the Planck’s constant, $f$ is the frequency and $W_0$ is the work function.

This is a straight line equation with $h$ being the slope and $-W_0$ being the $y$-intercept value.

At cutoff frequency, $f_0 = 1.0 times 10^{15}:Hz$, the kinetic energy of the electron is zero.

So the work function can be written as

$W_0 = hf_0 = 6.626 times 10^{-34}:J. cancel{s} times 1.0 times 10^{15} :dfrac{1}{cancel{s}} = 6.626 times 10^{-19} :J$