Question

If sodium is irradiated with light of 439nm, determine the maximum possible kinetic energy of the emitted electrons in eV

Answer 1

Photon is called the packet of quanta. It is basically an elementary particle. It is the basic unit of light. From high to low energy levels when an electron falls it emits a photon.

To calculate photon’s energy formula used is:

E = hv

where E is energy

h is Planck's constant

v is frequency

$h = 6.626 \times 10^{-34} \;J$

c = speed of light $= 3 \times 10^8 \; m/s$

Wavelength

$λ = 439 \;nm = 4.39 \times 10^{-7} \; m \\ v= \frac{c}{λ}$

c = speed of light $= 3 \times 10^8 \; m/s$

$v= \frac{3 \times 10^8}{4.39 \times 10^{-7}} = 0.683 \times 10^{15} \\ E = 6.626 \times 10^{-34} \times 0.683 \times 10^{15} \\ = 4.525 \times 10^{-19} \; J$

Sodium metal requires a photon with a minimum energy of $4.41 \times 10^{-19} \; J$ to emit electrons.

The maximum kinetic energy for each freed electron will be;

$4.525 \times 10^{-19} - 4.41 \times 10^{-19} \\ = 1.15 \times 10^{-20} \;J \\ = 1.15 \times 10^{-20} \times 6.242 \times 10^{18} = 0.0717 \;eV$