Question

When light with a wavelength of 238 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.13 10-19 J. Determine the wavelength of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.

Answer 1

Answer:

The wavelength is 173 nm.

Explanation:

This kind of phenomenon is known as photoelectric effect, it occurs when photons of light inside the metal surface and if they have the right amount of energy electrons absorb it and got expelled from the metal as photo electrons. The maximum kinetic energy of that photo electrons is given by the expression:

$K_{max} =E_{photon} - \Phi$ (1)

With E the energy of the photon and Φ the work function of the material. The work function is a value characteristic of each material and is related with how much the electron is attached to the material, the energy of the photon is the Planck's constant (h=$6.63\times10^{-34}$) times the frequency of light ($\nu$) , then (1) is:

$K_{max} =h\nu - \Phi$ (2)

The frequency of an electromagnetic wave is related with the wavelength ($\lambda$) by:

$\nu=\frac{c}{\lambda}$ (3)

with c the velocity of light (c=$3.0\times10^{8}$)

Using (3) on (2):

$K_{max} =\frac{hc}{\lambda} - \Phi$

Solving for $\Phi$:

$\Phi=\frac{hc}{\lambda}-K_max=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{238\times10^{-9}}-3.13\times10^{-19}$

$\Phi=5.23\times10^{-19} J$

That's the work function of the metal we're dealing. So now if we want to know the wavelength to obtain the double of the kinetic energy we use:

$2K_{max} =\frac{hc}{\lambda} - \Phi$

Solving for $\lambda$:

$\lambda = \frac{hc}{2K_{max}+\Phi}=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{2(3.13\times10^{-19})+5.23\times10^{-19}}=1.73\times10^{-7}$

$\lambda=173 nm$