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A certain metal when irradiated to light `(v = 3.2 xx 10^(16) Hz)` emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiation by light `( v = 2.0 xx 10^(16)Hz)`. The `v_(0)` Threshold frequency ) of the metal isA. `1.2xx10^(14)`HzB. `8xx10^(15)` HzC. `1.2xx10^(16)` HzD. `4xx10^(12)` Hz |
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Answer» Correct Answer - B `(K.E)_(1)=hv_(1)-hv_(0) (K.E)_(2)=hv_(2)-hv_(0) As" "(K.E)_(1)=2xx(K.E)_(2) therefore" "(hv_(1)-hv_(0))=2(hv_(2)-hv_(0)) or" "v_(0)=2v_(2)-v_(1)=2xx(2xx10^(16))-(3.2xx10^(16)) =0.8xx10^(16)Hz or 8xx10^(15)Hz` |
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