When photons of energy 3.5 eV fall on a metal surface, the stopping potential is 1.2 V. If photons of energy 5.0 eV fall on the same surface, the stopping potential will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2.7 V
The stopping potential measures the maximum kinetic energy of the photoelectrons, \( eV_0=K_{max} \). By Einstein’s photoelectric equation, \( eV_0=E-\phi_0 \), where \( E \) is the photon energy. The work function belongs to the surface, so it is the same in both cases. It is found from the first measurement and then used for the second.
From the first measurement, the work function is:
\( \begin{aligned} \phi_0 &= E_1-eV_{01} \\ &= 3.5\,\mathrm{eV}-1.2\,\mathrm{eV} \\ &= 2.3\,\mathrm{eV} \end{aligned} \)
With photons of energy 5.0 eV:
\( \begin{aligned} eV_{02} &= E_2-\phi_0 \\ &= 5.0\,\mathrm{eV}-2.3\,\mathrm{eV} \\ &= 2.7\,\mathrm{eV} \end{aligned} \)
Therefore, \( V_{02}=2.7\,\mathrm{V} \). The stopping potential grows by the same 1.5 V as the photon energy grows by 1.5 eV. It does not grow in proportion to the photon energy, because the same work function is taken away each time.