Work function for caesium metal is 2.14 eV. Let a beam of light of frequency `6xx10^(14)Hz` incident over the metal surface.
Now, match the following columns and choose the correct option from code given.

Options

A. A-4,B-3,C-2,D-1
B. A-3,B-4,C-2,D-1
C. A-3,B-1,C-4,D-2
D. A-2,B-1,C-4,D-3

Answer

B

Explanation

`(Ato3,Bto4,Cto2,Dto1)`
Maximum KE is given by
`K_(max)=hf-phi_(0)=(6.62xx10^(-34)xx6xx10^(14))/(1.6xx10^(-19))-2.14`
`=2.485-2.140=0.345`eV (so, `Ato3`)
Minimum kinetic energy,
`K_("min")=0` (so `Bto4`)
Stopping potential, `V_(0)=(K_(max))/(e)=(0.345eV)/(e)`
or `V_(0)=0.345V=345mV` (so, `Cto2`)
Maximum speed of emitted electrons,
`K_(max)=(1)/(2)_(max)^(2) or v_(max)=sqrt((2K_(max))/(m))=sqrt(0.1104xx10^(12))`
`=3.323xx10^(5)ms^(-1)=332kms^(-1)` (so, `Dto1`)