Zinc has a work function of 4.30 ev .

A 4.50eV photon strikes the surface and an electron is emitted. What is the maximum possible speed of the electron?

convert .2ev to joules, then

set it equal to 1/2 mv^2, and solve for v

To find the maximum possible speed of the emitted electron, we can use the concept of conservation of energy. The work function is the minimum energy required to remove an electron from a solid's surface. When a photon strikes the surface, it transfers its energy to the electron, causing it to be emitted.

The maximum kinetic energy of the emitted electron can be given by the difference between the energy of the incident photon and the work function:

Kinetic energy = Energy of incident photon - Work function

In this case, the energy of the incident photon is given as 4.50 eV (electron volts) and the work function is 4.30 eV.

Kinetic energy = 4.50 eV - 4.30 eV
Kinetic energy = 0.20 eV

Now, to convert the kinetic energy from electron volts to joules, we'll use the conversion factor: 1 eV = 1.6 × 10^-19 J.

Kinetic energy (in joules) = 0.20 eV × 1.6 × 10^-19 J/eV
Kinetic energy (in joules) = 3.2 × 10^-20 J

Finally, we can find the maximum possible speed of the electron using the kinetic energy:

Kinetic energy = (1/2)mv^2 (where m is the mass and v is the velocity)

Since we are given the mass of the electron (9.11 × 10^-31 kg), we can rearrange the equation:

v^2 = (2 × Kinetic energy) / m
v = √[(2 × Kinetic energy) / m]

Plugging in the values:

v = √[(2 × 3.2 × 10^-20 J) / (9.11 × 10^-31 kg)]

Now, calculate the maximum possible speed of the electron using this equation.