You have three metals - lithium, aluminum, and mercury. Their work functions are: 2.30, 4.10eV, and 4.5eV respectively. Light with a frequency of 1 x 10^15 Hz is incident on all three metals. Determine (a) which metals will emit electrons and (b)the maximum kinetic energy for those that exhibit the effect.

(a) The photon energy is

E = h*f = 6.62*10^-34 * 1*10^15 = 6.62*10^-19 J
Since 1 eV = 1.602*10^-19 J,
E = 4.13 eV.
Lithium and aluminum have work functions lower than that, and will emit photoelectrons

(b) The maximum photoelectron kinetic energy equals the photon energy (in eV) minus the work function

To determine which metals will emit electrons and the maximum kinetic energy for those metals, we need to compare the energy of the incident light to the work function of each metal.

(a) To find which metals will emit electrons, we compare the energy of the incident light with the work function of each metal. The energy of a photon can be calculated using the equation:

E = hf

where E is the energy, h is Planck's constant (6.63 x 10^-34 J·s), and f is the frequency of the light.

Given that the frequency of the light is 1 x 10^15 Hz, we can calculate the energy of the incident light as follows:

E = (6.63 x 10^-34 J·s) * (1 x 10^15 Hz)
Note: Hz stands for hertz, which represents cycles per second.

Solving the above equation, we find that E = 6.63 x 10^-19 J.

Now, we can compare the energy of the incident light to the work functions of each metal.

- For lithium: The work function of lithium is 2.30 eV. We need to convert this to joules using the conversion factor 1 eV = 1.6 x 10^-19 J.

2.30 eV * (1.6 x 10^-19 J/eV) = 3.68 x 10^-19 J

Since the energy of the incident light (6.63 x 10^-19 J) is greater than the work function of lithium (3.68 x 10^-19 J), lithium will emit electrons.

- For aluminum: The work function of aluminum is 4.10 eV.

4.10 eV * (1.6 x 10^-19 J/eV) = 6.56 x 10^-19 J

Since the energy of the incident light (6.63 x 10^-19 J) is greater than the work function of aluminum (6.56 x 10^-19 J), aluminum will also emit electrons.

- For mercury: The work function of mercury is 4.5 eV.

4.5 eV * (1.6 x 10^-19 J/eV) = 7.20 x 10^-19 J

Since the energy of the incident light (6.63 x 10^-19 J) is less than the work function of mercury (7.20 x 10^-19 J), mercury will not emit electrons.

Therefore, lithium and aluminum will emit electrons when the incident light with a frequency of 1 x 10^15 Hz is incident on them.

(b) The maximum kinetic energy of the emitted electrons can be determined using the equation:

Kinetic energy = Energy of incident light - Work function

For lithium:
Kinetic energy = 6.63 x 10^-19 J - 3.68 x 10^-19 J
Kinetic energy = 2.95 x 10^-19 J

For aluminum:
Kinetic energy = 6.63 x 10^-19 J - 6.56 x 10^-19 J
Kinetic energy = 7 x 10^-22 J

Therefore, the maximum kinetic energy for lithium is 2.95 x 10^-19 J, and for aluminum, it is 7 x 10^-22 J.