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(B) The Brackett series corresponds to electronic transitions ending at the $n = 4$ energy level.The first line of the Brackett series occurs for the

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$13.06$

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(B) The Brackett series corresponds to electronic transitions ending at the $n = 4$ energy level.The first line of the Brackett series occurs for the transition from $n = 5$ to $n = 4$.To emit this line,the hydrogen atom must first be excited from the ground state $(n = 1)$ to the $n = 5$ state.The energy required for this excitation is given by $\Delta E = E_5 - E_1$.Using the formula $E_n = -13.6 / n^2 \ eV$,we have:$E_1 = -13.6 \ eV$$E_5 = -13.6 / 5^2 = -13.6 / 25 = -0.544 \ eV$.Therefore,$\Delta E = -0.544 - (-13.6) = 13.056 \ eV \approx 13.06 \ eV$.Thus,the potential required to accelerate the electron is $13.06 \ V$,which corresponds to an energy of $13.06 \ eV$.

When a hydrogen sample in the ground state is bombarded with electrons,what potential is required to accelerate the electrons so that the first Brackett line is emitted? $(eV)$

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