The stopping potential for the photoelectrons | vedclass

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Question

$(A)$ According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by $K_{max} = e V_{0} = E - \p

Vedclass · Accepted Answer

$n = 3$ to $n = 1$

Vedclass · Answer

$(A)$ According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by $K_{max} = e V_{0} = E - \phi_{0}$, where $E$ is the energy of the incident photon, $V_{0}$ is the stopping potential, and $\phi_{0}$ is the work function.Given: $V_{0} = 10.4 \ V$ and $\phi_{0} = 1.7 \ eV$.Thus, $E = e V_{0} + \phi_{0} = 10.4 \ eV + 1.7 \ eV = 12.1 \ eV$.The energy of a photon emitted during a transition in a hydrogen atom from energy level $n_{i}$ to $n_{f}$ is given by $\Delta E = 13.6 \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} ight) \ eV$.For transition from $n = 3$ to $n = 1$:$\Delta E = 13.6 \left( \frac{1}{1^{2}} - \frac{1}{3^{2}} ight) = 13.6 \left( 1 - \frac{1}{9} ight) = 13.6 	imes \frac{8}{9} \approx 12.09 \ eV \approx 12.1 \ eV$.Therefore, the transition from $n = 3$ to $n = 1$ corresponds to the energy of the incident photons.

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