Question

Ask Your Teacher An experimenter finds that no photoelectrons are emitted from a particular metal unless the wavelength of light is less than 225 nm. Her experiment will require photoelectrons of maximum kinetic energy 2.2 eV. What frequency light should be used to illuminate the metal?

Answer 1

`\\u=5.31*10^(14) Hz`

This is green light

Step-by-step explanation:

According to the photoelectric effect, the maximum kinetic energy of an photoelectron is given by:

`K_(max)=E-W(1)`

Here, E is the energy of the photon and W is the minimum energy required to remove an electron from the surface of the metal, W is defined as:

`W=h\\u_0(2)`

The Planck – Einstein relation states that the energy of a photon is equal to its frequency multiplied by the planck constant:

`E=h\\u(3)`

Recall that
`\\u=(c)/(\lambda)`. Replacing (3) and (2) in (1):

`K_(max)=h\\u-h\\u_0\\K_(max)=h\\u-(hc)/(\lambda_0)`

Solving for
`\\u`:

`\\u=(K_(max)+(hc)/(\lambda_0))/(h)\\\\2.2eV*(1.60*10^(-19)J)/(1ev)=3.52*10^(-19)J\\\\\\u=(3.52*10^(-19)J+(6.63*10^(-34)J(3*10^8(m)/(s)))/(225*10^(-9)m))/(6.63*10^(-34)J)\\\\\\u=5.31*10^(14) Hz`

This is green light