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The work function `(phi)` of some metals is listed below. The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal is `{:("Metal",Li,Na,K,Mg,Cu,Ag,Fe,Pt,W),(phi(eV),2.4,2.3,2.2,3.7,4.8,4.3,4.7,6.3,4.75):}` |
Answer» Correct Answer - 4 Energy associated with the incident photon `= (hc)/(lamda)` i.e., `E = ((6.6 xx 10^(-34) Js) (3 xx 10^(8) ms^(-1)))/((300 xx 19^(-9) m))` `= 6.6 xx 10^(-19) J = (6.6 xx 10^(-19))/(1.6 xx 10^(-19)) eV = 4.12 eV` `:.` Metals showing photoelectric effect will be Li, Na, K and Mg only i.e., 4 metals, (which have work function less than 4.12 eV) |
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