Question: A photon with a frequency of 6.0 x 10^15 Hz strikes a metal surface, causing photoelectric emission. The work function of the metal is 4.0 eV. Determine the kinetic energy and velocity of the emitted electron.
Answer:
Given: Frequency of the photon (ν) = 6.0 x 10^15 Hz Work function of the metal (ϕ) = 4.0 eV
We can start by converting the frequency of the photon to energy using the Planck's equation:
E = hν
Where: E is the energy of the photon h is the Planck's constant (6.63 x 10^(-34) J s) ν is the frequency of the photon
Converting the frequency to energy: E = (6.63 x 10^(-34) J s) * (6.0 x 10^15 Hz) E = 3.978 x 10^(-18) J
Now, we need to convert the work function of the metal from electron volts (eV) to joules (J) using the conversion factor:
1 eV = 1.602 x 10^(-19) J
Converting the work function to joules: ϕ = (4.0 eV) * (1.602 x 10^(-19) J/eV) ϕ = 6.408 x 10^(-19) J
Since the energy of the photon (3.978 x 10^(-18) J) is greater than the work function (6.408 x 10^(-19) J), the excess energy will be converted into kinetic energy of the emitted electron.
The kinetic energy of the emitted electron can be calculated as:
K.E. = E - ϕ
K.E. = (3.978 x 10^(-18) J) - (6.408 x 10^(-19) J) K.E. = 3.337 x 10^(-18) J
Finally, we can determine the velocity of the emitted electron using the equation for kinetic energy:
K.E. = (1/2)mv^2
Where: K.E. is the kinetic energy of the electron m is the mass of the electron v is the velocity of the electron
Rearranging the equation to solve for velocity: v = √(2K.E. / m)
The mass of an electron is approximately 9.11 x 10^(-31) kg.
Plugging in the values: v = √[(2 * 3.337 x 10^(-18) J) / (9.11 x 10^(-31) kg)] v = 2.332 x 10^6 m/s
Therefore, the kinetic energy of the emitted electron is 3.337 x 10^(-18) J, and the velocity of the emitted electron is 2.332 x 10^6 m/s.
Note: The answer for velocity is given in scientific notation, where "2.332 x 10^6 m/s" means "2.332 multiplied by 10 to the power of 6 m/s".