What are the conditions for electron emission in a photocell, and how can the threshold frequency and Planck's constant be determined from a KEmax vs. frequency graph?

Here are the solutions to the questions:

a)
A condition necessary for electrons to be emitted from the cathode of a photocell is that the frequency of the incident light must be greater than or equal to the threshold frequency of the cathode material. Alternatively, the energy of the incident photons must be greater than or equal to the work function of the cathode material.

b)
The graph shows the relationship between the maximum kinetic energy ($KE_{max}$) of photoelectrons and the frequency ($f$) of incident light. According to the photoelectric equation, $KE_{max} = hf - \phi$, where $h$ is Planck's constant and $\phi$ is the work function.

i) Threshold frequency, $f_0$
The threshold frequency is the minimum frequency at which electrons are emitted, meaning $KE_{max} = 0$. On the graph, this is the x-intercept where the line crosses the frequency axis.

From the graph, the line intersects the x-axis at $f = 0.5$.
The x-axis is labeled $f (\times 10^{14} \text{ Hz})$.
Therefore, the threshold frequency is:

$$f_0 = 0.5 \times 10^{14} \text{ Hz}$$ $$f_0 = \boxed{\text{5.0} \times 10^{13} \text{ Hz}}$$

ii) Planck's constant, $h$
Planck's constant ($h$) is the gradient (slope) of the $KE_{max}$ versus $f$ graph.
We can choose two points from the line to calculate the gradient.
Let's use the points $(f_1, KE_{max1})$ and $(f_2, KE_{max2})$.

From the graph:
Point 1: $(0.5 \times 10^{14} \text{ Hz}, 0 \text{ J})$
Point 2: $(2.0 \times 10^{14} \text{ Hz}, 10.0 \times 10^{-19} \text{ J})$

Calculate the gradient:

$$h = \frac{\Delta KE_{max}}{\Delta f} = \frac{KE_{max2} - KE_{max1}}{f_2 - f_1}$$ $$h = \frac{(10.0 \times 10^{-19} \text{ J}) - (0 \text{ J})}{(2.0 \times 10^{14} \text{ Hz}) - (0.5 \times 10^{14} \text{ Hz})}$$ $$h = \frac{10.0 \times 10^{-19} \text{ J}}{1.5 \times 10^{14} \text{ Hz}}$$ $$h = 6.666... \times 10^{-33} \text{ J s}$$

Rounding to three significant figures:

$$h = \boxed{\text{6.67} \times 10^{-33} \text{ J s}}$$