Question

If the work function for a certain metal is 1.8eV, what is the stopping potential for electrons ejected from the metal when light of wavelength 400nm shines on the metal? And what is the maximum speed of the ejected electrons?

Answer 1

Just want to know if i'm going correct with the question..
\[E = E _{k} + \phi \]
\[E _{k} = eV _{0}\]
and to find speed you could use the equation for kinetic energy?
just want to make sure is stopping potential unitless??

Answer 2

You are indeed on the correct track. Just remember that the energy of a photon of wavelength lambda is given by \[E=\frac{hc}{\lambda}\]
you can then calculate the maximum kinetic energy since you know the work function of the metal (1.8eV), and the wavelength of the photons (400 nm) (remembering to convert to SI units of joules for the work function). you can then find the maximum velocity of the electrons easy enough.

Stopping potential (or stopping voltage) is then obtained by \[E_{k}=eV_{0}\]Because it is a stopping voltage, V_{0} will have the units of volts.

Answer 3

So to find speed of the ejected electrons would be found using this equation?? \[V _{0} = hc/e \lambda\] but why can't we use the \[E = 1/2mv ^{2}\] to find the speed?? is it because the speed of the electrons and photons depends on the frequency rather than the mass of the particle?? just want to make sure as i have an exam tomorrow :/ thanks for the help.

Answer 4

V_{0} is the stopping voltage for electrons ejected by a photon of wavelength lambda. The kinetic energy with which they are ejected will be the difference between the work function and the energy of the photons. So to find the maximum speed that they are ejected at (for a particular wavelength), the equation is\[\frac{1}{2}mv^{2}=\frac{hc}{\lambda}-\phi\] and solve for v.

But you are correct, the speed of the electrons depends upon the frequency (and hence wavelength) of the photons, as well as the work function of the metal.

To clarify one extra point (hopefully), the stopping voltage is the voltage required to stop electrons being emitted, i.e. the potential energy of the applied electric field to stop the electrons (eV_{0}) must equal the kinetic energy of the electrons, hence \[eV_{0}=\left(\frac{1}{2}mv^{2}\right)_{max}\]