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A photon of energy `8 eV` is incident on a metal surface of threshold frequency `1.6 xx 10^(15) Hz`, then the maximum kinetic energy of photoelectrons emitted is `( h = 6.6 xx 10^(-34) Js)`A. `4.8 eV`B. `2.4 eV`C. `1.4 eV`D. `0.8 eV` |
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Answer» Correct Answer - C Work function `W_(0) = hv_(0) = 6.6 xx 10^(-34) xx 1.6 xx 10^(15)` `= 1.056 xx 10^(-18) J = 6.6 eV` From `E = W_(0) + K_(max) rArr K_(max) = E - W_(0) = 1.4 eV` |
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