Question

A metal surface is illuminated with light of different wavelengths. It is observed that electrons are emitted from the metal for wavelengths of light up to 525 nm but for no wavelengths above 525 nm. When light of 420 nm is used, what is the maximum kinetic energy of the electrons?

Answer 1

`K.E.=9.4657* 10^(-20)\ J`

Step-by-step explanation:

Using the expression for the photoelectric effect as:

`E=h\\u_0+\frac {1}{2}* m* v^2`

Also,
`E=\frac {h* c}{\lambda}`

`\\u_0=\frac {c}{\lambda_0}`

Applying the equation as:

`\frac {h* c}{\lambda}=\frac {hc}{\lambda_0}+\frac {1}{2}* m* v^2`

Where,

h is Plank's constant having value
`6.626* 10^(-34)\ Js`

c is the speed of light having value
`3* 10^8\ m/s`

`\lambda` is the wavelength of the light being bombarded

`\lambda_0` is the threshold wavelength

`\frac {1}{2}* m* v^2` is the kinetic energy of the electron emitted.

Given,
`\lambda=420\ nm=420* 10^(-9)\ m`

`\lambda_0=525\ nm=525* 10^(-9)\ m`

Thus, applying values as:

`\frac {6.626* 10^(-34)* 3* 10^8}{420* 10^(-9)}=\frac {6.626* 10^(-34)* 3* 10^8}{525* 10^(-9)}+K.E.`

`K.E.=\frac {6.626* 10^(-34)* 3* 10^8}{420* 10^(-9)}-\frac {6.626* 10^(-34)* 3* 10^8}{525* 10^(-9)}`

`K.E.=(19.878)/(10^(17)* \:420)-(19.878)/(10^(17)* \:525)`

`K.E.=9.4657* 10^(-20)\ J`