Question

Calculate the threshold energy in kj/mol of electrons in aluminum, given that the lowest frequency photon for which the photoelectric effect is observed is 9.87 × 1014 hz.

Answer 1

Answer: The threshold energy of the electrons in aluminium is 393.71 kJ/mol.

Explanation: Threshold energy is the minimum amount of energy which is required to emit a photon from a metal surface.

Mathematically,

$E=N_Ah\nu_o$

Where,

E = Threshold energy

$N_A$ = Avogadro's constant = $6.022\times 10^{23}atoms/mol$

h = Planck's Constant = $6.624\times 10^{-34}Js$

$\nu$ = Frequency of the particle =$9.87\times 10^{14}s^{-1}$

Putting values in above equation, we get:

$E=(6.022\times 10^{23})atoms/mol\times (6.624\times 10^{-34})Js\times (9.87\times 10^{14})s^{-1}$

$E=393.71\times 10^3J/mol=393.71kJ/mol$   (Conversion Factor: 1kJ = 1000J)