A photon with an energy of 2.5 eV interacts with an electron in a metal surface and is completely absorbed. The electron is initially at rest and is emitted with a kinetic energy of 1.5 eV. Determine:
a) The work function of the metal. b) The speed of the emitted electron.
Answer with Step-by-Step Detailed Explanation:
a) The work function refers to the minimum energy required to remove an electron from the surface of a material.
We know that the energy of a photon is given by:
E = hf
where E is the energy of the photon, h is Planck's constant (6.63 x 10^(-34) J s), and f is the frequency of the photon.
Since the problem provides the energy of the photon as 2.5 eV, let's convert it to joules using the conversion factor 1 eV = 1.6 x 10^(-19) J:
E = (2.5 eV) * (1.6 x 10^(-19) J/eV) E = 4.0 x 10^(-19) J
Now, we need to find the work function, which is the minimum energy required to remove an electron. In this case, the work function can be determined by subtracting the kinetic energy of the emitted electron from the energy of the photon:
Work Function = Energy of Photon - Kinetic Energy of Electron Work Function = (4.0 x 10^(-19) J) - (1.5 eV) * (1.6 x 10^(-19) J/eV) Work Function = 0.1 x 10^(-19) J
Therefore, the work function of the metal is 0.1 x 10^(-19) J.
b) To find the speed of the emitted electron, we can use the principle of energy conservation.
The kinetic energy of the emitted electron can be calculated using the equation:
KE = (1/2) mv^2
where KE is the kinetic energy, m is the mass of the electron (9.11 x 10^(-31) kg), and v is the speed of the electron.
Since the kinetic energy of the emitted electron is given as 1.5 eV, let's convert it to joules using the conversion factor 1 eV = 1.6 x 10^(-19) J:
KE = (1.5 eV) * (1.6 x 10^(-19) J/eV) KE = 2.4 x 10^(-19) J
Now, we can set the kinetic energy equal to the energy difference between the photon and the work function:
KE = Energy of Photon - Work Function 2.4 x 10^(-19) J = (4.0 x 10^(-19) J) - (0.1 x 10^(-19) J)
Rearranging the equation, we can solve for the speed of the electron:
v^2 = (2 * (2.4 x 10^(-19) J)) / m v^2 = 1.64 x 10^7 m^2/s^2
Taking the square root of both sides, we can find the speed of the emitted electron:
v = sqrt(1.64 x 10^7 m^2/s^2)
Hence, the speed of the emitted electron is approximately 4.05 x 10^3 m/s.