Light of wavelength 200 nm falls on an aluminum surface, having workfunction 4.2 eV. What is the K.E. of the fastest emitted photoelectron? Alsofind the stopping potential and cutoff wavelength.

Sure, let's solve this step by step.

1. First, we need to convert the given wavelength of light into energy using the formula E = hc/λ, where h is Planck's constant (6.626 x 10^-34 Js), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength. But before we do that, we need to convert the wavelength from nm to m. So, 200 nm = 200 x 10^-9 m.

2. Now, we can calculate the energy of the incident light: E = (6.626 x 10^-34 Js x 3 x 10^8 m/s) / (200 x 10^-9 m) = 9.939 x 10^-19 J.

3. To make the next steps easier, let's convert this energy to electron volts (eV). 1 eV = 1.6 x 10^-19 J, so E = 9.939 x 10^-19 J / 1.6 x 10^-19 J/eV = 6.21 eV.

4. The work function of aluminum is given as 4.2 eV. This is the minimum energy required to remove an electron from the surface of the aluminum.

5. The kinetic energy (K.E.) of the fastest emitted photoelectron can be found using the photoelectric effect equation: K.E. = E - work function. Substituting the given values, we get K.E. = 6.21 eV - 4.2 eV = 2.01 eV.

6. The stopping potential (V) can be found using the equation V = K.E./e, where e is the charge of an electron (1.6 x 10^-19 C). Substituting the values, we get V = 2.01 eV / 1.6 x 10^-19 C = 1.26 V.

7. The cutoff wavelength (λc) can be found using the equation λc = hc / work function. Substituting the values, we get λc = (6.626 x 10^-34 Js x 3 x 10^8 m/s) / (4.2 eV x 1.6 x 10^-19 J/eV) = 296.7 nm.

So, the kinetic energy of the fastest emitted photoelectron is 2.01 eV, the stopping potential is 1.26 V, and the cutoff wavelength is 296.7 nm.