Solve the following showing all the steps, in English

The quantum energy $E$ of radiation is given by Planck's relation:

$$E = hf$$

where $h$ is Planck's constant and $f$ is the frequency of the radiation.

We are given $h = 6.63 \times 10^{-34}$ J s and $f = 2.5 \times 10^{14}$ Hz.

$$E = (6.63 \times 10^{-34} \text{ J s}) \times (2.5 \times 10^{14} \text{ Hz})$$

$$E = 16.575 \times 10^{-20} \text{ J}$$ $$E = 1.6575 \times 10^{-19} \text{ J}$$

The quantum energy of the radiation is $\boxed{\text{1.6575} \times 10^{-19} \text{ J}}$.

The work function $W_0$ is given as 3.0 eV.
The conversion factor is 1 eV $= 1.6 \times 10^{-19}$ J.

$$W_0 = 3.0 \text{ eV} \times \left(1.6 \times 10^{-19} \frac{\text{J}}{\text{eV}}\right)$$

$$W_0 = 4.8 \times 10^{-19} \text{ J}$$

The work function in Joules is $\boxed{\text{4.8} \times 10^{-19} \text{ J}}$.

The wavelength $\lambda = 3.5$ nm.
We know that 1 nm $= 10^{-9}$ m.
So, $\lambda = 3.5 \times 10^{-9}$ m.
Planck's constant $h = 6.64 \times 10^{-34}$ J s (for this part).
The speed of light $c = 3.0 \times 10^8$ m/s.
The energy of the incident photon $E$ is given by:

$$E = \frac{hc}{\lambda}$$ $$E = \frac{(6.64 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m/s})}{3.5 \times 10^{-9} \text{ m}}$$

$$E = \frac{19.92 \times 10^{-26}}{3.5 \times 10^{-9}} \text{ J}$$ $$E \approx 5.6914 \times 10^{-17} \text{ J}$$

The photoelectric effect equation is $E = W_0 + K_{max}$, where $K_{max}$ is the maximum kinetic energy.
Therefore, $K_{max} = E - W_0$.
We have $W_0 = 4.8 \times 10^{-19}$ J (calculated previously).

$$K_{max} = (5.6914 \times 10^{-17} \text{ J}) - (4.8 \times 10^{-19} \text{ J})$$

To subtract, we need to have the same power of 10:

$$K_{max} = (569.14 \times 10^{-19} \text{ J}) - (4.8 \times 10^{-19} \text{ J})$$ $$K_{max} = (569.14 - 4.8) \times 10^{-19} \text{ J}$$ $$K_{max} = 564.34 \times 10^{-19} \text{ J}$$ $$K_{max} = 5.6434 \times 10^{-17} \text{ J}$$

The maximum kinetic energy of the photoelectrons is $\boxed{\text{5.6434} \times 10^{-17} \text{ J}}$.

The threshold frequency $f_0$ is the minimum frequency required to eject an electron, and it is related to the work function by $W_0 = hf_0$.
Therefore, $f_0 = \frac{W_0}{h}$.
We use $W_0 = 4.8 \times 10^{-19}$ J and $h = 6.64 \times 10^{-34}$ J s.

$$f_0 = \frac{4.8 \times 10^{-19} \text{ J}}{6.64 \times 10^{-34} \text{ J s}}$$

$$f_0 \approx 0.72289 \times 10^{15} \text{ Hz}$$ $$f_0 \approx 7.2289 \times 10^{14} \text{ Hz}$$

The threshold frequency is $\boxed{\text{7.2289} \times 10^{14} \text{ Hz}}$.