Question

In a photoelectric-effect experiment, the maximum kinetic energy of electrons is 3.0 eV . When the wavelength of the light is increased by 50%, the maximum energy decreases to 1.2 eV . What is the work function of the cathode? What is the initial wavelength?

Answer 1

2.4 eV

230 nm

Step-by-step explanation:

The equation for the photoelectric effect is:

`Emax = h * f - \phi`

Where

h: Planck's constant (4.13e-15 eV * s)

f: frequency

phi: work function

This can be omdified to:

`Emax = h * (c)/(\lambda) - \phi`

Where

c: speed of light in vacuum

lambda: wave length

We can set two equations:

(1)
`3 = h * (c)/(\lambda) - \phi`

(2)
`1.2 = h * (c)/(1.5 * \lambda) - \phi`

If we multiply equation (2) by 1.5 we obtain

(3)
`1.8 = h * (c)/(1.5 * \lambda) - 1.5 * \phi`

If we substract eq (3) from eq (1)

`1.2 = 0.5 * \phi`

`\phi =(1.2)/(0.5) = 2.4eV`

Knowing this we can calculate the original wavelength

`\lambda = h * (c)/(3 + \phi)`

`\lambda = 4.13e-15 * (3e8)/(3 + 2.4) = 2.3e-7 m = 230 nm`