Question

Question #0e47b

Answer 1

`2.89 * 10^4"kJ mol"^(-1)`

Explanation:

The idea here is that the incoming photon will transmit its energy to the electron.

If this energy is high enough to overcome the binding energy of the electron, the electron will be emitted from the surface with a kinetic energy `K_E`.

Now, the difference between the energy of the incoming photon and the kinetic energy of the emitted electron will be equal to the binding energy.

`color(blue)(E_"binding" = E_"photon" - K_E`

The energy of the photon is directly proportional to its frequency, `nu`

`color(blue)(E = h * nu)`

Frequency is inversely proportional to wavelength, `lamda`. This means that you can write

`color(blue)( nu * lamda = c)" "`

and therefore

`color(blue)(E = (h * c)/(lamda))" "`, where

`h` - Planck's constant, equal to `6.626 * 10^(-34)"J s"`
`c` - the speed of light in a vacuum, usually given as `3 * 10^8 "m s"^(-1)`

Use this equation to find the energy of the incoming photon - do not forget to convert the wavelength from nanometers to meters!

`E = (6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 3 * 10^8 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(0.990 * 10^(-9)color(red)(cancel(color(black)("m"))))`

`E = 2.01 * 10^(-16)"J"`

Next, focus on converting the kinetic energy of the electron from electron-volts to joules

`953 color(red)(cancel(color(black)("eV"))) * (1.602 * 10^(-19)"J")/(1color(red)(cancel(color(black)("eV")))) = 1.53 * 10^(-16)"J"`

This means that the binding energy for one electron will be

`E_"binding" = 2.01 * 10^(-16)"J" - 1.53 * 10^(-16)"J"`

`E_"binding" = 4.80 * 10^(-17)"J"`

To get the binding energy for one mole of electrons, use the Avogadro's number, which tells you that one mole contains exactly `6.022 * 10^(23)` "things" in it.

In this case, one mole of electrons will contain `6.022 * 10^(23)` electrons.

`4.80 * 10^(-17)"J"/color(red)(cancel(color(black)("e"^(-)))) * overbrace( (6.022 * 10^(23)color(red)(cancel(color(black)("e"^(-)))))/"1 mole e"^(-))^(color(purple)("Avogadro's number")) = 2.89 * 10^7"J mol"^(-1)`

Finally, convert this from joules per mole to kilojoules per mole

`2.89 * 10^(7)color(red)(cancel(color(black)("J")))/"mol" * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = color(green)(2.89 * 10^(4)"kJ mol"^(-1))`

The answer is rounded to three sig figs.