Question

An arsenic surface emits electrons after shining with a light source of the frequency of 5.32×1023 Hz. Calculate the maximum kinetic energy of the ejected electron if the work function of arsenic is 7.67×10.

Answer 1

The maximum kinetic energy of the ejected electron=
`3.53* 10^(-10) J`

Step-by-step explanation:

We are given that

Frequency of light source=
`5.32* 10^(23)` Hz

Work function of arsenic=
`7.67* 10` eV

We have to find the maximum kinetic energy of ejected electron.

We know that the maximum kinetic energy of ejected electron

`K.E=h\\u-w_o`

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

`\\u` =Frequency of light source

`w_o`=Work function

Substitute the values in the given formula

Then, the maximum kinetic energy of ejected electron

`K.E=6.63* 10^(-34)* 5.32* 10^(23)-7.67* 10* 1.6* 10^(-19)`

Because 1 e V=
`1.6* 10^(-19) J`

`K.E=35.2716* 10^(-11)-12.272* 10^(-18)`

`K.E=3.53* 10^(-10)` J

Hence, the maximum kinetic energy of the ejected electron=
`3.53* 10^(-10) J`