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If light of frequency 8.2xx10^14 Hz is i...

If light of frequency `8.2xx10^14` Hz is incident on the metal , cut-off voltage for photoelectric emission is 2.03 V.Find the threshold frequency . Given h=`6.63xx10^(-34) ` Js , e=`1.6xx10^(-19)` C.

Text Solution

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`hv=hv_0+exx2.03`
`v_0=(hv-exx2.03)/h`
`=8.2xx10^14 -(1.6xx10^(-19)xx2.03)/(6.63xx10^(-34))`
`=(8.2xx4.9)xx10^14`
`3.3xx10^14`Hz
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