Question

The following table lists the work functions of a few commonmetals, measured in electron volts.

Answer 1

Answer:

Lithium

Explanation:

The equation for the photoelectric effect is

$\frac{hc}{\lambda}= \phi + K_{max}$

where

$\frac{hc}{\lambda}$ is the energy of the incident photon, with

h being the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

$\phi$ is the work function of the metal (the minimum energy needed to extract the photoelectron from the metal)

$K_{max}$ is the maximum kinetic energy of the emitted photoelectrons

In this problem, we have

$\lambda= 190 nm = 1.9\cdot 10^{-7}m$ is the wavelength of the incident photon

$K_{max}=4.0 eV$ is the maximum kinetic energy of the electrons

First of all we can find the energy of the incident photon

$E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.90\cdot 10^{-7} m}=1.05\cdot 10^{-18} J$

Converting into electronvolts,

$E=\frac{1.05\cdot 10^{-18} J}{1.6\cdot 10^{-19} J/eV}=6.6 eV$

So now we can re-arrange the equation of the photoelectric effect to find the work function of the metal

$\phi = E-K_{max}=6.6 eV - 4.0 eV=2.6 eV$

So the metal is most likely Lithium, which has a work function of 2.5 eV.