A beam of light with a frequency of 7.61•1015 Hz hits the surface of the metal. What is the velocity of the electron ejected in m/s? (6 pts)

How do you find the velocity based only off the frequency?

To determine the velocity of the ejected electron based only on the frequency of the incident light, you can use the concept of the photoelectric effect. The photoelectric effect states that when light, in the form of photons, hits a metal surface, it can cause electrons to be ejected, provided that each photon has enough energy to overcome the binding energy of the electron in the metal.

The energy of a photon (E_photon) can be calculated using the formula:

E_photon = h * f

Where h is Planck's constant (6.62607004 × 10^-34 J·s) and f is the frequency of the light.

Once you know the energy of the photon, you can equate it to the kinetic energy (KE) of the ejected electron using the equation:

KE = (1/2) * m * v^2

Where m is the mass of the electron (9.10938356 × 10^-31 kg) and v is the velocity of the electron.

Since the photon's energy is equal to the electron's kinetic energy, we can write:

h * f = (1/2) * m * v^2

Rearranging the equation to solve for the velocity (v), we get:

v = sqrt((2 * h * f) / m)

Now, substituting the given values:
h = 6.62607004 × 10^-34 J·s
f = 7.61 × 10^15 Hz
m = 9.10938356 × 10^-31 kg

We plug these values into the formula and calculate the velocity:

v = sqrt((2 * 6.62607004 × 10^-34 J·s * 7.61 × 10^15 Hz) / 9.10938356 × 10^-31 kg)

v ≈ 6.57 × 10^5 m/s

Therefore, the velocity of the ejected electron is approximately 6.57 × 10^5 m/s.