Answer:
The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm
Explanation:
Given;
wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m
maximum kinetic energy of the ejected electron, K.E = 0.92 eV
let the work function of the aluminum metal = Ф
Apply photoelectric equation:
E = K.E + Ф
Where;
Ф is the minimum energy needed to eject electron the aluminum metal
E is the energy of the incident light
The energy of the incident light is calculated as follows;
![E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J](https://tex.z-dn.net/?f=E%20%3D%20hf%20%3D%20h%5Cfrac%7Bc%7D%7B%5Clambda%7D%20%5C%5C%5C%5Cwhere%3B%5C%5C%5C%5Ch%20%5C%20is%20%5C%20Planck%27s%20%5C%20constant%20%3D%206.626%20%5Ctimes%2010%5E%7B-34%7D%20%5C%20Js%5C%5C%5C%5Cc%20%5C%20is%20%5C%20speed%20%5C%20of%20%5C%20light%20%3D%203%20%5Ctimes%2010%5E%7B8%7D%20%5C%20m%2Fs%5C%5C%5C%5CE%20%3D%20%5Cfrac%7B%286.626%5Ctimes%2010%5E%7B-34%7D%29%5Ctimes%20%283%5Ctimes%2010%5E8%29%7D%7B248%5Ctimes%2010%5E%7B-9%7D%7D%20%5C%5C%5C%5CE%20%3D%208.02%20%5Ctimes%2010%5E%7B-19%7D%20%5C%20J)
The work function of the aluminum metal is calculated as;
Ф = E - K.E
Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)
Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J
Ф = 6.546 x 10⁻¹⁹ J
The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;