Question

When light of wavelength 345 nm falls on a potassium surface, electrons are emitted that have a maximum kinetic energy of 1.67 eV. What is the work function of potassium? The speed of light is 3 × 108 m/s and Planck’s constant is 6.63 × 10−34 J · s.

Answer 1

`W=1.93eV`

Step-by-step explanation:

The maximum kinetic energy of an ejected electron in the photoelectric effect is given by:

`K_(max)=h\\u-W(1)`

Here h is the Planck's constant,
`\\u` the frequency of the light and W the work function of the element.

The frequency is equal to the speed of light, divided by the wavelength:

`\\u=(c)/(\lambda)(2)`

Recall that
`1nm=10^(-9)m`. Replacing (2) in (1) and solving for W:

`W=(hc)/(\lambda)-K_(max)\\W=((4.14*10^(-15)eV\cdot s)(3*10^8(m)/(s)))/(345*10^(-9)m)-1.67eV\\W=1.93eV`