Question

When a surface is illuminated with electromagnetic radiation of wavelength 480 nm, the maximum kinetic energy of the emitted electrons is 0.54 eV. What is the maximum kinetic energy if the surface is illuminated using radiation of wavelength 340 nm?

Answer 1

Answer:

Max kinetic energy for 340 nm wavelength will be $2.238\times 10^{-19}j$

Explanation:

In first case wavelength of electromagnetic radiation $\lambda =480nm=480\times 10^{-9}m$

Plank's constant $h=6.6\times 10^{-34}J-s$

Maximum kinetic energy = 0.54 eV

Energy is given by $E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{480\times 10^{-9}}=4.125\times 10^{-19}J$

We know that energy is given

$E=K_{MAX}+\Phi$, here $\Phi$ is work function

So $4.125\times 10^{-19}=0.54\times 10^{-19}+\Phi$

$\Phi =3.585\times 10^{-19}J$

Now wavelength of second radiation = 340 nm

So energy $E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{340\times 10^{-9}}=5.823\times 10^{-19}J$

So $K_{MAX}=5.823\times 10^{-19}-3.585\times 10^{-19}=2.238\times 10^{-19}j$