The work function of the nickel is 5 eV. When a light of wavelength 2000 Å falls on it ,it emits photoelectrons in the circuit .Then, the potential difference necessary to stop the faster electrons emitted is (Given ,`h=6.67xx10^(-34)J-s)`

Options

A. 1.0V
B. 1.75V
C. 1.5V
D. 0.75V

Answer

C

Explanation

Work function m`phi=5eV=5xx1.6xx10^(-1)`
Wavelength ,`lamda`=2000Å
The expression for stopping potential is given as `eV_(o)(hc)/(lamda)-phiimplies eV_(o)=((6.67xx10^(-34))(3xx10^(8)))/(2xx10^(-7))=-(5xx1.6xx10^(-19)`
`implies eV_(o)=(10.005xx10^(-19))-(8xx10^(19))=2.005xx10^(-19)`
`implies eV_(o)=(10.005xx10^(-19))-(8xx10^(19))=2.005xx10^(-19)`
`implies V_(o)=(2.005xx10^(-19))/(1.6xx10^(-19))=1.25V`