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

This sounds like it's about the photoelectric effect.

The binding energy

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)#

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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{

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*Answer from HIX Tutor*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

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