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A metal surface is illuminated by light ...

A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are `u_1` and `u_2` respectively. If the ratio `u_1: u_2 = 2:1 and hc= 1240 eV` nm, the work function of the metal is neraly. (a)3.7 eV (b) 3.2 eV (c ) 2.8eV (d) 2.5eV.

A

`3.7 eV`

B

`3.2 eV`

C

`2.8 eV`

D

`2.8 eV`

Text Solution

Verified by Experts

The correct Answer is:
A
`248 nm 1240//248 eV =5eV`
`310 nm 1250//310 eV = 4eV`
`(KE_(1))/(KE_(2))=(4)/(1) =(5ev-W)/(4ev-W)`
`rArr 16-4 W =S-W rArr 11 = 3W`
`rArr W = (11)/(3)=3.67 ev 3.7 ev`
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