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

Answer:

$W=1.93eV$

Explanation:

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

$K_{max}=h\nu-W(1)$

Here h is the Planck's constant, $\nu$ 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:

$\nu=\frac{c}{\lambda}(2)$

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

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