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When a 6000Å light falls on the cathode ...

When a `6000Å` light falls on the cathode of a photo cell and produced photoemission. If a stopping potential of 0.8 V is required to stop emission of electron, then determine the
threshold frequency

Text Solution

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Given data `:`
Wavelength of incident light `lambda =6000Å = 6 xx 10^(-7) m `
Stopping potential `V_(0) =0.8 V`
To find `:`
(i) Frequency of light v = ?
(ii) Energy of incident photon E =?
(iii) Work function of the cathode material W= ?
(iv) Threshold frequency `v_(0) = ?`
(v)Net energy of the electron after it leaves the surface E = ?
Threshold frequency `= v_(0) =( phi_(0))/(h)`
`= ( 2.01 xx 10^(-19))/( 6.626 xx 10^(-34)) = 0.303 xx 10^(15)`
`v_(0) = 3.03 xx 10^(14) Hz`
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