According to Bohr's atomic model,which of the | vedclass

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(C) The energy of the $n^{	ext{th}}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \; eV$.The energy levels are:$E_1 = -1

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$11.1$

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(C) The energy of the $n^{	ext{th}}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \; eV$.The energy levels are:$E_1 = -13.6 \; eV$$E_2 = -\frac{13.6}{4} = -3.4 \; eV$$E_3 = -\frac{13.6}{9} \approx -1.51 \; eV$$E_4 = -\frac{13.6}{16} = -0.85 \; eV$The energy of an emitted photon is the difference between two energy levels: $\Delta E = E_{n_2} - E_{n_1}$.Checking the options:$1$. $E_4 - E_3 = -0.85 - (-1.51) = 0.66 \; eV$ (approx $0.65 \; eV$). This is possible.$2$. $E_3 - E_2 = -1.51 - (-3.4) = 1.89 \; eV$ (approx $1.9 \; eV$). This is possible.$3$. $E_1$ is the ground state energy,and transitions to $E_1$ can release $13.6 \; eV$ (e.g.,from infinity to $n=1$). This is possible.$4$. $11.1 \; eV$ cannot be expressed as the difference between any two energy levels of the hydrogen atom.Therefore,$11.1 \; eV$ is not a possible energy for an emitted photon.

According to Bohr's atomic model,which of the following is $NOT$ a possible energy for a photon emitted by a hydrogen atom (in $;eV$)? (in $eV$)

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