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The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photo electrons was found to be 1.68 eV. The work function of the metal is `(h_(c)=1240eV.nm)`

A

1.41 eV

B

1.51eV

C

1.68eV

D

3.09 eV

Text Solution

Verified by Experts

The correct Answer is:
A

Here `lamda=400nm and K_(max)=1.68eV. " "therefore` Energy of incident light photon `E=(hc)/(lamda)=(1240)/(400)eV=3.1eV`
As `E=phi_(0)+K_(max),` hence `phi_(0)=E-K_(max)=3.1-1.68=1.42eV`
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