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

Answer:

2.4 eV

230 nm

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:

$Emax = h * \frac{c}{\lambda} - \phi$

Where

c: speed of light in vacuum

lambda: wave length

We can set two equations:

(1) $3 = h * \frac{c}{\lambda} - \phi$

(2) $1.2 = h * \frac{c}{1.5 * \lambda} - \phi$

If we multiply equation (2) by 1.5 we obtain

(3) $1.8 = h * \frac{c}{1.5 * \lambda} - 1.5 * \phi$

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

$1.2 = 0.5 * \phi$

$\phi =\frac{1.2}{0.5} = 2.4eV$

Knowing this we can calculate the original wavelength

$\lambda = h * \frac{c}{3 + \phi}$

$\lambda = 4.13e-15 * \frac{3e8}{3 + 2.4} = 2.3e-7 m = 230 nm$