by how much would the stopping potential for a given  photosensitive surface go up if the frequency of the incident radiations were to be increased from 4*10^15 hz to 8 *10^15 hz​

Dear Student ,
Here in this case stopping potential Vs is given by ,
eVs=hν-WVs=heν-WeWhen , ν1=4×1015 Hz , Vs=Vs1sayWhen , ν2=8×1015 Hz , Vs=Vs2sayTherefore,Vs1=heν1-We       and   Vs2=heν2-WeSubtracting we get , Vs2- Vs1=heν2-ν1=6·4×10-341·6×10-198×1015-4×1015=16 volt .
Regards

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