Question

 Calculate the maximum kinetic energy of electrons emitted from photosensitive surface of work function 3.2 eV, for the incident radiation of wavelength 300 nm.

Answer 1

hf = K + W
K = hf - W
K = hc/λ - W
K = (1240 eV nm)/(300 nm) - 3.2 eV
K = 4.13eV - 3.2eV
K = 0.93 eV
K = 0.93 * 1.6 * 10^-19 J
K = 1.488 * 10^-19 J

Answer 2

Given:

• The incident radiation = 300 nm = 300 × $10^9$ m
• Work function = 3.2eV

To Find:

• The maximum Kinetic energy of the electrons.

Solution:

We have a formula saying,

⇒ $E_{in} = w_0+E_{max}$ → {equation 1}

Where $E_{in}$ is the energy of the incident photon, $w_0$ is the work function, $E_{max}$ is the maximum kinetic energy.

First, we need to find the value  $E_{in}$ by using a formula,

⇒ $E_{in}$ = hc/λ → {equation 2}

Where "h" is Planck's constant ( 6.6×$10^{-34}$), "c" is the speed of light ( 3×$10^8$), "λ" is the wavelength.

Substitute the values in equation 2. We get,

⇒ $E_{in}$ = (6.6×$10^{-34}$ × 3×$10^8$)/(300×$10^{-9}$)  J  

We should convert the SI unit of the above equation to eV as the other values are in the same SI unit.

$E_{in}$  =  (6.6×$10^{-34}$ × 3×$10^8$)/(300×$10^{-9}$×1.6×$10^{-19}$) eV

⇒ $E_{in}$ = 4.125 eV

Now substitute the value of $E_{in}$ and $w_0$ in equation 1. We get,

⇒ 4.125 = 3.2+$E_{max}$

rearrange the above equation to find the value of $E_{max}$

⇒ $E_{max}$ = 4.125-3.2 = 0.925 eV

∴ The maximum kinetic energy = 0.925 eV.