Next, suppose that the beam of electrons is shut off so that all of the objects are in the ground state almost all the time. if electromagnetic radiation with a wide range of energies is passed through the material, what will be the three energies of photons corresponding to missing ("dark") lines in the spectrum? remember that there is hardly any absorption from excited states, because emission from an excited state happens very quickly, so there is never a significant number of objects in an excited state. assume that the detector is sensitive to a wide range of photon energies, not just energies in the visible region. list the dark-line energies in order from largest to smallest.
a. ev (largest)b. evc. ev (smallest)

A. 3.9 eV (largest) B. 3.4 eV  C. 2.0 eV (smallest)

Explanation:

Let's call the excited states 1, 2, and 3. The ground state is state 0. Assuming that the initial state is the highest excited state i.e. state 3, the emission of photon can occur via transitions to other excited states and energy will be lost by the object. The same thing will occur for the other excited states i.e. states 2 and 1. On the contrary, no emission occur for the state 0 (i.e. the ground state).

Therefore, we will subtract the ground state (0) from each excited state (i.e. 1, 2, 3). Thus:

 -0.7 eV - (-4.6 eV) = 3.9 eV

-1.2 eV -(-4.6 eV) = 3.4 eV

-2.6 eV – (4.6 eV) = 2.0 eV

Therefore,  A. 3.9 eV (largest) B. 3.4 eV  C. 2.0 eV (smallest).

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.