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Surface of certain metal is first illumi...

Surface of certain metal is first illuminated with light of wavelength `lambda_(1)=350` nm and then, by light of wavelength `lambda_(2)=540` nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to :
(Energy of photon `=(1240)/(lambda("in nm"))Ev`

A

2.5

B

1.8

C

5.6

D

1.4

Text Solution

Verified by Experts

The correct Answer is:
B

`1240/540 -phi=kimplies2.296-phi=k`……..i
`1240/350-phi=4kimplies3.543-phi=4k`….ii
Solving I and ii `k=k0.416, phi=1.88eV`
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