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Electrons with a kinetici energy of 6.02...

Electrons with a kinetici energy of `6.023xx10^(4)J//mol` are evolved from the surface of a metal, when it is exposed to radiation of wavelength of 600 nm. The minimum amount of energy required to remove an electron fro the metal atom is:

A

`2.3125xx10^(-19)J`

B

`3xx10^(-19)J`

C

`6.02xx10^(-19)J`

D

`6.62xx10^(-34)J`

Text Solution

Verified by Experts

The correct Answer is:
A
Maximum kinetic energy of ejected electron = absorbed energy - threshold energy
Kinetic energy of 1 mol of electron = `6.022xx10^(4)J`
Kinetic energy of 1 electron = `(6.022xx10^(4))/(6.022xx10^(23))JrArr10^(-19)J`
So, `K.E.=10^(-19)J=hv-hv_(0)`
`10^(-19)=(hc)/lamda-"threshold energy"`
Threshold energy = `(6.62xx10^(-34)xx3xx10^(8))/(600xx10^(-9))-10^(-19)`
= `3.31xx10^(-19)-10^(-19)J`
= `2.31xx10^(-19)J`.
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