CBSE Class 12-science Answered
the stopping potential in an experiment of photon is 2v what is the maximum kinetic energy of photoelectrons emitted?
Asked by vasturushi | 28 Feb, 2018, 12:49: PM
Expert Answer
I assume the stopping potential given in the question is 2 eV
The relation connecting the stopping potential and maximum kinetic energy is given by
Kmax = eVo ; Kmax is the maximum kinetic energy, Vo is stopping potential and e is electronic charge.
Kmax = 2 eV = 2 × 1.602 × 10-19 Joules = 3.2 × 10-19 Joules
Answered by Thiyagarajan K | 28 Feb, 2018, 03:29: PM
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