Question

If the threshold wavelength for copper is 2665 A, calculate the maximum kinetic energy of a photoelectron generated by ultraviolet light of λ 2000 A. Compute the stopping potential.

Answer 1

`K.E.=2.48* 10^(-19)\ 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=2000\ \dot{A}=2000* 10^(-10)\ m`

`\lambda_0=2665\ \dot{A}=2665* 10^(-10)\ m`

Thus, applying values as:

`\frac {6.626* 10^(-34)* 3* 10^8}{2000* 10^(-10)}=\frac {6.626* 10^(-34)* 3* 10^8}{2665* 10^(-10)}+K.E.`

`K.E.=\frac {6.626* 10^(-34)* 3* 10^8}{2000* 10^(-10)}-\frac {6.626* 10^(-34)* 3* 10^8}{2665* 10^(-10)}`

`K.E.==(19.878)/(10^(16)* \:2000)-(19.878)/(10^(16)* \:2665)`

`K.E.=2.48* 10^(-19)\ J`