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.
A
`1.6xx10^(18)Hz`
B
`0.8xx10^(15)Hz`
C
`8xx10^(15)Hz`
D
`8xx10^(16)Hz`
Text Solution
Verified by Experts
The correct Answer is:
C
When a certain metal was ………. `hv = hv_(0)+KE` `v-v_(0)=KE//h` `v_(2)-v_(0)=2v_(1)-v_(2)` `= 4xx10^(16) - 3.2xx10^(18)` `= 0.8xx10^(16) = 8xx10^(15)Hz`
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