What frequency of light is needed to eject an electron from a metal whose threshold energy is #"201 kJ/mol"#?

Question

Wyatt Adkins · Accepted Answer

#5.04 xx 10^14# #"Hz"# or #"s"^(-1)#

This sounds like it's about the photoelectric effect.

The binding energy #phi# is the energy that would have needed to be supplied by an incoming light source in order to remove a desired number of electrons from the surface (in this case, #"1 mol"# of electrons).

It is also defined as the threshold that needs to be overcome in order to eject a specific number of electron(s). Hence, its associated frequency IS the threshold frequency.

If you supply more energy than the required threshold to eject a specific quantity of electrons, the difference in the supplied energy and the binding energy gives the remaining kinetic energy that goes into the ejected electrons.

Therefore, the equation becomes:

#bb(stackrel("leftover kinetic energy")overbrace(K_e) = stackrel("supplied")overbrace(E_"photon"))# #-bb()# #bb(stackrel("required threshold")overbrace(phi))#
#= bb(hnu - hnu_0)#
where #nu_0# is the threshold frequency.
Based on your question, all you need is the threshold frequency #nu_0#, which means:

#phi = hnu_0#
#color(blue)(nu_0) = phi/h#
#=# #[(201 cancel"kJ")/cancel"mol electrons" xx cancel"1 mol electrons"/(6.022xx10^(23)) xx (1000 cancel"J")/cancel"1 kJ"]/[6.626xx10^(-34) cancel("J")cdot"s"]#
#= color(blue)(5.04 xx 10^14# #color(blue)("Hz")# or #"s"^(-1)#

Peyton Adams · Answer

undefined The frequency of light needed to eject an electron from a metal can be calculated using the equation $E = h \times f$, where $E$ is the energy required to eject the electron, $h$ is Planck's constant ($6.626 \times 10^{-34}$ J·s), and $f$ is the frequency of the light.

First, convert the threshold energy from kilojoules per mole to joules per atom:

$$\text{Threshold energy} = \frac{201 \text{ kJ/mol}}{6.022 \times 10^{23} \text{ atoms/mol}}$$

Then, solve for the frequency of light:

$$f = \frac{E}{h}$$

$$f = \frac{\text{Threshold energy}}{h}$$

$$f = \frac{201 \times 10^3 \text{ J/mol}}{6.022 \times 10^{23} \text{ atoms/mol} \times 6.626 \times 10^{-34} \text{ J·s}}$$

$$f ≈ \frac{3.34 \times 10^{-19}}{6.626 \times 10^{-34}} \text{ s}^{-1}$$

\[f ≈ 5.04 \times 10^{14} \text{