Question

When light with wavelength 640 nm shines on a particular sheet of metal, it ejects electrons that have a maximum kinetic energy of 0.3 eV . What is the maximum kinetic energy of ejected electrons if we shine light with wavelength 300 nm on the sheet of metal instead?

Answer 1

To solve this problem it is necessary to apply the concepts related to energy in photons. Although these although they have no mass have energy and are defined as

$E = hf= \frac{hc}{\lambda}$

Where,

c = Speed of light

h = Planck's constant$\rightarrow 6.626*10^{-34}$

$\lambda$ = wavelength

f= Frequency of the radiation

In addition to this we know that the maximum kinetic energy in a photon is given by the formula

$k_{max} = hf-\phi$

Where,

$\phi =$ Amount of energy binding the electron to the metal

$\phi$ can be expressed also as

$\phi = \frac{hc}{\lambda}-k$

Replacing the terms then we have to

$E = K+\phi$

$\frac{hc}{\lambda_2} = k_2 +(\frac{hc}{\lambda_1}-k_1)$

Re-arrange to find $k_2,$

$k_2 = \frac{hc}{\lambda_2} -(\frac{hc}{\lambda_1}-k_1)$

$k_2 = \frac{hc}{\lambda_2} -\frac{hc}{\lambda_1}+k_1$

Replacing with our values,

$k_2 = \frac{4.136*10^{-15}*3*10^17}{300} -\frac{4.136*10^{-15}*3*10^17}{640}$

$k_2 = 2.19725+0.3$

$k_2 = 2.49eV$

Therefore the maximum kinetic energy is 2.49eV

Note: Speed of light is converted to nanometers.

Answer 2

Answer:

The maximum kinetic energy is 2.5 eV

Explanation:

please look at the solution in the attached Word file