The work function of very pure Na is 2.75 eV, where 1 eV = 1.602×10−19 J.(a) Calculate the maximum kinetic energy of photoelectrons emitted from Na exposed to 260 nm light.(b) Calculate the longest wavelength that will initiate photoelectric effect in pure Na

(a) To calculate the maximum kinetic energy of photoelectrons emitted from Na exposed to 260 nm light, we first need to calculate the energy of the incident light. The energy of a photon is given by the equation E = hc/λ, where 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, convert the wavelength from nm to m: 260 nm = 260 x 10^-9 m.

Then, substitute these values into the equation:

E = (6.626 x 10^-34 J.s)(3 x 10^8 m/s) / (260 x 10^-9 m)

E = 7.65 x 10^-19 J

To convert this to eV, divide by the charge of an electron (1.602 x 10^-19 J/eV):

E = (7.65 x 10^-19 J) / (1.602 x 10^-19 J/eV)

E = 4.77 eV

The maximum kinetic energy of the photoelectrons is then given by the energy of the incident light minus the work function of the material. In this case, the work function of Na is given as 2.75 eV. So:

K.E_max = E - Work function

K.E_max = 4.77 eV - 2.75 eV

K.E_max = 2.02 eV

(b) To calculate the longest wavelength that will initiate the photoelectric effect in pure Na, we need to find the wavelength of light that has energy equal to the work function of Na. We can rearrange the equation E = hc/λ to solve for λ:

λ = hc / E

Substitute the work function of Na for E (remembering to convert it to J):

λ = (6.626 x 10^-34 J.s)(3 x 10^8 m/s) / (2.75 eV x 1.602 x 10^-19 J/eV)

λ = 4.49 x 10^-7 m

To convert this to nm, multiply by 10^9:

λ = 4.49 x 10^-7 m x 10^9 nm/m

λ = 449 nm

So, the longest wavelength that will initiate the photoelectric effect in pure Na is 449 nm.