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When a certain metal was irradiated with light of frequency `3.2 xx 10^(16)s^(-1)` the photoelectrons emitted had twice the KE as did photoelectrons emitted when the same metal was irradiated with light of frequency`2.0 xx 10^(16)s^(-1)` . Calculate the thereshold frequency of the metal. |
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Answer» Correct Answer - `319.2kJ//mol` `hv_(1)=hv_(0)+2E_(1)" "hv_(2)=hv_(0)+E_(1)` `hv_(1)-w_(0)+2E_(1)" "hv_(2)-w_(0)+E_(1)` `2=(hv_(1)-w_(0))/(hv_(2)-w_(0))" "2 hv_(2)-2w_(0)=hv_(1)-w_(0)` `h[2v_(2)-v_(1)]=w_(0)` `w_(0)=6.62xx10^(-34)(2xx10^(15)-3.2xx10^(15))` `w_(0)=6.62xx10^(-34)xx0.8xx10^(15)` `w_(0)=5.29xx10^(-19)" "w_(0)=318.9kJ//mol` |
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