Question

When a metal was exposed to light at a frequency of 3.64× 1015 s–1, electrons were emitted with a kinetic energy of 5.80× 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 8.66× 10–7 J?

Answer 1

Answer : The maximum number of electrons released
`4.73* 10^(12)electrons`

Explanation : Given,

Frequency =
`3.64* 10^(15)s^(-1)`

Kinetic energy =
`5.80* 10^(-19)J`

Total energy =
`8.66* 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.

`5.80* 10^(-19)J=(6.626* 10^(-34)J/s* 3.64* 10^(15)s^(-1))-w`

By rearranging the terms, we get

`w=1.83* 10^(-18)J`

Therefore, the works function of the metal is,
`1.83* 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}}`

`(8.66* 10^(-7)J)/(1.83* 10^(-19)J)=4.73* 10^(12)electrons`

Therefore, the maximum number of electrons released is
`4.73* 10^(12)electrons`