Question

A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second light source of wavelength l/2 ejects photoelectrons with a maximum kinetic energy of 4 eV. What is the work function of the metal

Answer 1

Step-by-step explanation:

From first source, kinetic energy (
`K.E_(1)`) ejected is 1 eV and wavelength of light is
`\lambda`.

From second source, kinetic energy (
`K.E_(2)`) ejected is 4 eV and wavelength of light is
`(\lambda)/(2)`.

Relation between work function, wavelength, and kinetic energy is as follows.

K.E =
`(hc)/(\lambda) - \phi`

where, h = Plank's constant =
`6.63 * 10^(-34)` J.s

c = speed of light =
`3 * 10^(8)` m/s

Also, it is known that 1 eV =
`1.6 * 10^(-19)` J

Therefore, substituting the values in the above formula as follows.

• From first source,

`K.E_(1)` =
`(hc)/(\lambda) - \phi`

1 eV =
`1.6 * 10^(-19) J = (6.63 * 10^(-34) * 3 * 10^(8))/(\lambda) - \phi`

`1.6 * 10^(-19) J = (1.98 * 10^(-25) J.m)/(\lambda) - \phi` ........... (1)

• From second source,

`K.E_(2)` =
`(hc)/(\lambda) - \phi`

`4 * 1.6 * 10^(-19) J = (6.63 * 10^(-34) * 3 * 10^(8))/((\lambda)/(2)) - \phi`

`6.4 * 10^(-19) J = (2 * 1.98 * 10^(-25) J.m)/(\lambda) - \phi` ........... (2)

Now, divide equation (2) by 2. Therefore, it will become

`{6.4 * 10^(-19)J}{2} = (2 * 1.98 * 10^(-25) J.m)/(2\lambda) - (\phi)/(2)`

`3.2 * 10^(-19)J = (1.98 * 10^(-25) J.m)/(\lambda) - (\phi)/(2)` ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

`1.6 * 10^(-19) = (\phi)/(2)`

`\phi` =
`3.2 * 10^(-19)`

= 2 eV

Thus, we can conclude that work function of the metal is 2 eV.