Question

Question #31613

Answer 1

`1.8 * 10^(-19)"J"`

Explanation:

What you basically need to do here is use the equation that establishes a relationship between the energy of the incomining photon, the work function of the metal, and the kinetic energy of the ejected electron

`color(blue)(K_E = h * nu - W)" "`, where

`K_E` - the kinetic energy of the electron;
`h` - Planck's constant, equal to `6.626 * 10^(-34)"J" * "s"`
`nu` - the frequency of the incoming photon;
`W` - the work function of the metal.

The idea here is that in order for an electron to be ejected from the surface of the metal, the frequency of the incoming photon must be high enough so that the energy of the photon, `h * nu`, exceeeds thw work function.

Use the given wavelength to determine what the frequency of the photons is - here `c` represents the speed of light in vacuum

`color(blue)(lamda * nu = c implies nu = c/(lamda))`

This means that you have

`nu = (3 * 10^8color(red)(cancel(color(black)("m"))) * "s"^(-1))/(220 * 10^(-9)color(red)(cancel(color(black)("m")))) = 1.36 * 10^(15) "s"^(-1)`

Now, in order to determine the kinetic energy of a single electron, use Avogadro's number to calculate the work function per photon. It will be helpful along the way to convert from kJ to J

`435color(red)(cancel(color(black)("kJ")))/color(red)(cancel(color(black)("mol"))) * "1000 J"/(1color(red)(cancel(color(black)("kJ")))) * (1color(red)(cancel(color(black)("mole"))))/(6.022 * 10^(23)"photons") = 7.22 * 10^(-19)"J"`

Now plug these values into the main equation to get

`K_E = 6.626 * 10^(-34)"J" * color(red)(cancel(color(black)("s"))) * 1.36 * 10^(15)color(red)(cancel(color(black)("s"^(-1)))) - 7.22 * 10^(-19)"J"`

`K_E = 9.01 * 10^(-19)"J" - 7.22 * 10^(-19)"J"`

`K_E = color(green)(1.8 * 10^(-19)"J")`

The answer is rounded to two sig figs, the number of sig figs you gave for the wavelength of the photons.