Quantum Physics and photoelectrons

[WAHEEDA]

here are another questions,I need someone to help me!
(1)A metal work a work function of 2.40 eV is illuminated with a beam of monochromatic light.The stopping potential for the emmited electrons is 2.50 V.What is the wavelength of the light in nanometer?
(2)What is the lowest frequency of light that can cause the release of electrons from a metal that has a work function of 2.80 eV?
(3)Gold has a work function of 4.82 eV.If a block of gold is illuminated with ultraviolet light (wavelength=160nm), what are
(a) the maximum kinetic energy (in electron volts) of emmited photoelectrons
(b) the threshold frequency?

please help me!I need the answers by this monday,29th november

 

[Nate810]

(1) The wavelength of the light in nanometers is 1240. (2) The lowest frequency of light that can cause the release of electrons from a metal with a work function of 2.80 eV is 4.48 x 10^15 Hz. (3a) The maximum kinetic energy (in electron volts) of emmited photoelectrons is 2.62 eV. (3b) The threshold frequency is 7.59 x 10^14 Hz.

 

[wais]

(1) To find the wavelength of the light, we can use the equation: E = hc/λ, where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J*s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength of the light.

First, we need to find the energy of the photon using the given work function and stopping potential. The energy of the photon can be calculated by subtracting the stopping potential from the work function: E = 2.50 V - 2.40 eV = 0.10 eV.

Next, we need to convert the energy from electron volts (eV) to joules (J) by multiplying by the conversion factor 1.602 x 10^-19 J/eV. So, 0.10 eV = 0.10 x 1.602 x 10^-19 J = 1.602 x 10^-20 J.

Now, we can plug in the values into the equation: 1.602 x 10^-20 J = (6.626 x 10^-34 J*s)(3 x 10^8 m/s)/λ

Solving for λ, we get: λ = 6.626 x 10^-34 J*s x 3 x 10^8 m/s / 1.602 x 10^-20 J = 3.91 x 10^-7 m = 391 nm.

Therefore, the wavelength of the light is 391 nm.

(2) The lowest frequency of light that can cause the release of electrons from a metal with a work function of 2.80 eV can be found using the same equation as above, but this time we are solving for the frequency (f) instead of the wavelength (λ). The equation for frequency is: E = hf. Rearranging the equation, we get: f = E/h.

Substituting the values, we get: f = 2.80 eV / 6.626 x 10^-34 J*s = 4.23 x 10^14 Hz.

Therefore, the lowest frequency of light that can cause the release of electrons is 4.23 x 10^14 Hz.

(3) (a) To find the maximum kinetic energy of the emitted photoelectrons, we can use