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(B) The energy of the emitted photons from a hydrogen discharge tube corresponds to the ionization energy of hydrogen,which is $E = 13.6 \ eV$.Accordi

Vedclass · Accepted Answer

$-9.4$

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(B) The energy of the emitted photons from a hydrogen discharge tube corresponds to the ionization energy of hydrogen,which is $E = 13.6 \ eV$.According to Einstein's photoelectric equation,$E = W_0 + K_{max}$,where $K_{max} = e|V_0|$ and $V_0$ is the stopping potential.Substituting the given values: $13.6 \ eV = 4.2 \ eV + e|V_0|$.$e|V_0| = 13.6 \ eV - 4.2 \ eV = 9.4 \ eV$.Therefore,the magnitude of the stopping potential is $|V_0| = 9.4 \ V$.To reduce the photo-current to zero,the anode must be at a negative potential relative to the cathode.Thus,the required voltage is $-9.4 \ V$.

Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is $4.2 \ eV$. In order to reduce the photo-current to zero,the voltage of the anode relative to the cathode must be made ........... $V$.

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