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(B) The possible transitions for emission are from higher energy levels to lower energy levels. The energy of the emitted photon is given by $\Delta E

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$465$

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(B) The possible transitions for emission are from higher energy levels to lower energy levels. The energy of the emitted photon is given by $\Delta E = E_f - E_i$. The possible transitions are:$1$. From $E_3$ to $E_2$: $\Delta E_{32} = -2 \ eV - (-6 \ eV) = 4 \ eV$.$2$. From $E_3$ to $E_1$: $\Delta E_{31} = -2 \ eV - (-8 \ eV) = 6 \ eV$.$3$. From $E_2$ to $E_1$: $\Delta E_{21} = -6 \ eV - (-8 \ eV) = 2 \ eV$.Using the formula $\lambda = \frac{hc}{\Delta E}$,where $hc \approx 1240 \ eV \cdot nm$:- For $\Delta E = 4 \ eV$,$\lambda = \frac{1240}{4} = 310 \ nm$.- For $\Delta E = 6 \ eV$,$\lambda = \frac{1240}{6} \approx 206.67 \ nm \approx 207 \ nm$.- For $\Delta E = 2 \ eV$,$\lambda = \frac{1240}{2} = 620 \ nm$.Comparing these with the given options,$465 \ nm$ is not present in the emission spectrum.

$A$ hypothetical atom has only three energy levels. The ground level has energy,$E_1 = -8 \ eV$. The two excited states have energies,$E_2 = -6 \ eV$ and $E_3 = -2 \ eV$. Which of the following wavelengths will $NOT$ be present in the emission spectrum of this atom (in $nm$)?

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