Question

When a metal was exposed to light at a frequency of 4.07× 1015 s–1, electrons were emitted with a kinetic energy of 3.30× 10–19 J. What is the maximum number of electrons that could be ejected from this metal by a burst of light (at some other frequency) with a total energy of 3.39× 10–7 J?

Answer 1

Answer : The maximum number of electrons released =
`1.432* 10^(12)electrons`

Explanation : Given,

Frequency =
`4.07* 10^(15)s^(-1)`

Kinetic energy =
`3.30* 10^(-19)J`

Total energy =
`3.39* 10^(-7)J`

First we have to calculate the work function of the metal.

Formula used :

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

where,

K.E = kinetic energy

h = Planck's constant =
`6.626* 10^(-34)J/s`

`\\u` = frequency

w = work function

Now put all the given values in this formula, we get the work function of the metal.

`3.30* 10^(-19)J=(6.626* 10^(-34)J/s* 4.07* 10^(15)s^(-1))-w`

By rearranging the terms, we get

`w=2.367* 10^(-18)J`

Therefore, the works function of the metal is,
`2.367* 10^(-18)J`

Now we have to calculate the maximum number of electrons released.

The maximum number of electrons released =
`\frac{\text{ Total energy}}{\text{ work function}}`

The maximum number of electrons released =
`(3.39* 10^(-7)J)/(2.367* 10^(-19)J)=1.432* 10^(12)electrons`

Therefore, the maximum number of electrons released is
`1.432* 10^(12)electrons`