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Ultraviolet light of wavelength 66.26 nm...

Ultraviolet light of wavelength 66.26 nm and intensity `2 W//m^(2)` falls on potassium surface by which photoelectrons are ejected out. If only 0.1% of the incident photons produce photoelectrons, and surface area of metal surface is `4 m^(2)`, how many electrons are emitted per second?

A

`2.67 xx 10^(15)`

B

`3 xx 10^(15)`

C

`3.33 xx 10^(7)`

D

`4.17 xx 10^(16)`

Text Solution

Verified by Experts

The correct Answer is:
A
Number of photons falling on metal surface,
`n_(p)=("intensity"xx"area")/("energy per quanta")`
`=((2Js^(-1)m)xx(4m^(2)))/({{(6.626xx10^(-34)Js)xx(3xx10^(8)ms^(-1)))/((66.26xx10^(-9)m)}})`
`=2.67xx10^(18)" per sec"`
From equation, `n_(e)=0.1%" of "n_(p)`
`=(0.1)/(100)xx2.67xx10^(18)=2.67xx10^(15)" per sec"`
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