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(D) The work function of the cathode surface is $\phi = 4 \ eV$.In a hydrogen discharge tube,the maximum energy of a photon emitted during the transit

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$-10$

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(D) The work function of the cathode surface is $\phi = 4 \ eV$.In a hydrogen discharge tube,the maximum energy of a photon emitted during the transition of an electron from higher energy levels to the ground state $(n=1)$ is given by the Rydberg formula,where the maximum energy corresponds to the transition from $n = \infty$ to $n = 1$,which is $13.6 \ eV$.According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ of the emitted photoelectrons is given by $K_{max} = E_{photon} - \phi$.Substituting the maximum photon energy,$K_{max} = 13.6 \ eV - 4 \ eV = 9.6 \ eV$.To make the photocurrent zero,the stopping potential $(V_s)$ must be applied such that $e|V_s| = K_{max}$.Therefore,$|V_s| = 9.6 \ V$.Since the stopping potential must be negative with respect to the cathode to stop the electrons,the anode voltage must be at least $-9.6 \ V$. Among the given options,$-10 \ V$ is the only value that is more negative than $-9.6 \ V$,which will effectively stop all photoelectrons.

Light coming from a discharge tube filled with hydrogen falls on the cathode of the photoelectric cell. The work function of the surface of the cathode is $4 \ eV$. Which one of the following values of the anode voltage (in Volts) with respect to the cathode will likely make the photocurrent zero?

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