A certain metal was initiated with light of frequency 3.2×1016 Hz. The photoelectrons emitted had twice the kinetic energy as the same metal was irradiated with light of frequency 2×1016Hz. Calculate the threshold frequency.
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Solution
let,v1(nu)=3.2X10^16hz,v2(nu)=2X10^16hz acc. to ques,KE1=2xKE2 acc.to einsteis eq. KE=hv-hvo so hv1-hvo=2(hv2-hvo)or v1-vo=2(v2-vo) putting values 3.2X10^16-vo=2(2X10^16-vo) on solving u will get vo=0.8X10^16HZ ans.vo(thershold freq.)=0.8X10^16hz