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A radiation of 2000Å falls on the metal whose work function is 4.2 eV. Then the kinetic energy of the fastest photo eletron is |
Answer» <html><body><p>`6.4xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)` <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a><br/>`16xx10^(-10)` J<br/>`1.6xx10^(-19)` J<br/>`3.2xx10^(-19)` J</p>Solution :Kinetic energy (<a href="https://interviewquestions.tuteehub.com/tag/ke-527890" style="font-weight:bold;" target="_blank" title="Click to know more about KE">KE</a>) `=hv-W` <br/> `(<a href="https://interviewquestions.tuteehub.com/tag/hc-1016346" style="font-weight:bold;" target="_blank" title="Click to know more about HC">HC</a>)/lambda-4.2eV(6.62xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/34-308171" style="font-weight:bold;" target="_blank" title="Click to know more about 34">34</a>)xx3xx10^(8))/(2000xx10^(-10))=-4.2xx1.6xx10^(-19)` <br/> `=9.9xx10^(-19)j-6.7xx10^(-19)J=3.2xx10^(-19)` J</body></html> | |