Anode voltage is at +3 V. Incident radiation has frequency 1.4 × 105 Hz and work function of photocathode is 2.8 eV. Find the minimum and maximum kinetic energy of photoelectrons in ev
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Answer:3,6
Explanation:
KEmax=hv_Φ
=(6.625×10^_34×1.4×10^15/1.6×10^_19)_2.8
=3ev
Since anode voltage is +3 V, therefore an electron emitted with zero K.E. will
acquire an energy = 3 eV
The electron emitted with kinetic energy 3 eV will acquire energy
= 3eV + 3eV = 6 eV
Hence minimum KE = 3 eV and Max KE = 6 eV
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