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If light of wavelength 412.5 nm is incident on each of the metals given below, which one will show photoelectric emission and why ? |
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Answer» Solution :Calculating the energy of the incident phohton Identifying the metals Reason The energy of a photon of incident radiation is GIVEN by ` E= ( h_in ) /( lambda ) ` ` therefore E= ( 6.63xx 10 ^( -34) xx 3xx10 ^(8) ) /( ( 412.5xx 10 ^(-9)) xx (1.6xx10 ^(-19)) eV ` `~= 3.01eV ` Hence ,only Na and K will show photoelectric emission [Note : Award this mark even if the student wirtes the NAME of only one of these metals ] Reason : They energy of the incident photon is more that the work function of only these two metals. |
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