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The work function of a metal is 3.0 V. It is illuminated by a light of wave length 3xx10^(-7)m.Calculate i) threshold frequency, ii) the maximum energy of photoelectrons, iii) the stopping potential. (h=6.63xx10^(-34)Js and c=3xx10^(8)ms^(-)). |
| Answer» <html><body><p></p>Solution :`W=hv_(0)=3.0eV=3xx1.6xx10^(-19)J` <br/> <a href="https://interviewquestions.tuteehub.com/tag/threshold-1419373" style="font-weight:bold;" target="_blank" title="Click to know more about THRESHOLD">THRESHOLD</a> frequency <br/> `v_(0)=(W)/(<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>)=(3xx1.6xx10^(-19))/(6.63xx10^(-34))=0.72xx10^(15)Hz` <br/> (ii) Maximum kinetic energy `(E_(max))=h(v-v_(0))` <br/> `lamda-<a href="https://interviewquestions.tuteehub.com/tag/3xx10-1865443" style="font-weight:bold;" target="_blank" title="Click to know more about 3XX10">3XX10</a>^(-7)m,v=(c)/(lamda)=(3xx10^(8))/(3xx10^(-7))=1xx10^(15)Hz` <br/> `K_(max)=h(v-v_(0))` <br/> `=6.63xx10^(-34)(1-0.72)xx10^(15)J=1.86xx10^(-19)J`. <br/> (<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>) `K_(max)=eV_(0)` where `V_(0)` is stopping potential in volt and e is the charge of electron <br/> `V_(0)=(K_(max))/(e)`. Here `K_(max)=1.86xx10^(-19)J` and `e=1.6xx10^(-19)C" "V_(0)=(1.86xx10^(-19)J)/(1.6xx10^(-19)C)=1.16V`</body></html> | |