Question

cesium metal has a work function w = 3.12x10^(-19) j and can produce the photoelectric effect when irradiated with certain colours of visible light. if the maximum kinetic energy of an ejected electron is 1.06x10^(-19) j, what colour of light was used to produce the photoelectric effect?
a. blue (450 - 495 nm)
b. green (495 - 570 nm)
c. yellow (570 - 590 nm)
d. orange (590 - 620 nm)
e. red (620 - 750 nm)

Explanation:

Step1: Use the photoelectric - effect equation

The photoelectric - effect equation is $E_{k}=h
u - w$, where $E_{k}$ is the maximum kinetic energy of the ejected electron, $h
u$ is the energy of the incident photon, and $w$ is the work function of the metal. First, we need to find the energy of the incident photon. Rearranging the equation gives $h
u=E_{k}+w$.
$h
u=(1.06\times 10^{-19}\text{ J})+(3.12\times 10^{-19}\text{ J}) = 4.18\times 10^{-19}\text{ J}$

Step2: Use the energy - wavelength relationship

The energy of a photon is given by $E = h
u=\frac{hc}{\lambda}$, where $h = 6.63\times 10^{-34}\text{ J}\cdot\text{s}$, $c = 3\times 10^{8}\text{ m/s}$, and $\lambda$ is the wavelength of the light. Rearranging for $\lambda$ gives $\lambda=\frac{hc}{E}$.
$\lambda=\frac{(6.63\times 10^{-34}\text{ J}\cdot\text{s})\times(3\times 10^{8}\text{ m/s})}{4.18\times 10^{-19}\text{ J}}\approx476\text{ nm}$

Answer:

A. Blue (450 - 495 nm)