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Direction : The questions 64, 65 and 66 are based on the following paragraph: The work function of caesium metal is 2.14 eV. When light of frequency 6xx10^(14)Hz is incident, on the metal surface, the photoemission of electrons starts. Then stopping potential is

Answer»

0.349 V
0.67 V
0.81 V
1.2 V

Solution :STOPPING potential `(V_(S)):`
`eV_(s)=1//2 mv^(2) rArr V_(s)=(K.E.)/(e)`
`rArrV_(s)=(0.588xx10^(-19))/(1.6xx10^(-19))=0.349V`


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