Question

The energy of a photon needed to cause ejection of an electron from a photoemissive metal is expressed as the sum of the binding energy of the electron plus the kinetic energy of the emitted electron. When photons of 4.00×10-7 m light strike a calcium metal surface, electrons are ejected with a kinetic energy of 6.26×10-20 J. a Calculate the binding energy of the calcium electrons.

Answer 1

`Binding\ energy=43.43* 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

Given,
`\lambda=4.00* 10^(-7)\ m`

`\frac {hc}{\lambda_0}` is the binding energy or threshold energy

`\frac {1}{2}* m* v^2` is the kinetic energy of the electron emitted. =
`6.26* 10^(-20)\ J`

Thus, applying values as:

`(h* c)/(\lambda)=Binding\ Energy+Kinetic\ Energy`

`(6.626* 10^(-34)* 3* 10^8)/(4.00* 10^(-7))\ J=Binding\ Energy+6.26* 10^(-20)\ J`

`(19.878)/(10^(19)* \:4)\ J=Binding\ Energy+6.26* 10^(-20)\ J`

`49.69* 10^(-20)\ J=Binding\ Energy+6.26* 10^(-20)\ J`

`Binding\ energy=43.43* 10^(-20)\ J`