A stream of photons having energy 3 ,eV each | vedclass

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(A) The energy of the emergent photo-electron is given by Einstein's photoelectric equation: $E_{max} = h
u - \Phi_0$.Given,incident photon energy $h

Vedclass · Accepted Answer

$3.5$

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(A) The energy of the emergent photo-electron is given by Einstein's photoelectric equation: $E_{max} = h
u - \Phi_0$.Given,incident photon energy $h
u = 3 \,eV$ and work function $\Phi_0 = 2.3 \,eV$.Therefore,the maximum kinetic energy $E_{max} = 3 - 2.3 = 0.7 \,eV$.This implies that the electron will turn back when it encounters a retarding potential difference of $0.7 \,V$.The potential difference between the two plates is $1 \,V$ over a distance of $5 \,mm$.Assuming a uniform electric field,the potential gradient is $\frac{1 \,V}{5 \,mm} = 0.2 \,V/mm$.The distance $d$ required to reach a potential difference of $0.7 \,V$ is given by $d = \frac{V_{stop}}{	ext{gradient}} = \frac{0.7 \,V}{0.2 \,V/mm} = 3.5 \,mm$.Thus,the photo-electron turns back after traveling $3.5 \,mm$.

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