A hydrogen atom has its electron in the n $=3$ state. How much energy would have to be absorbed by the atom for it to become ionized from this level? What is the frequency of the photon that could produce this result?

An atom has $46$ electrons. What is the smallest value of n needed so that all $46$ electrons occupy the lowest possible quantum states consistent with the Pauli exclusion principle?

Group of answer choices

$1\mathrm{.}51$ eV $3\mathrm{.}65$ x $10^14$ Hz$,$ n $=4$

$1\mathrm{.}58$ eV $3\mathrm{.}93$ x $10^14$ Hz$,$ n $=5$

$1\mathrm{.}51$ eV $3\mathrm{.}65$ x $10^14$ Hz$,$ n $=5$

$1\mathrm{.}51$ eV $3\mathrm{.}93$ x $10^14$ Hz$,$ n $=5$

$1\mathrm{.}44$ eV $3\mathrm{.}65$ x $10^14$ Hz$,$ n $=6$