91Ó°ÊÓ

Whenever the bright lines fall on the metal surface, the emission of a photon takes place, and when the dark lines fall on the metal surface, the absorption of a photon takes place.

Given that the photon is getting absorbed with the energy of ${\mathrm{E}}_1−{\mathrm{E}}_0$.

For the absorption process the tail of the arrow should be in the initial state and the head of the arrow should be in the final state. As here energy is ${\mathrm{E}}_1−{\mathrm{E}}_0$, so the tail of the arrow should be at ${\mathrm{E}}_0$ and the head of the arrow should be at ${\mathrm{E}}_1$.

Therefore absorption of photon whose energy is${\mathrm{E}}_1−{\mathrm{E}}_0$ is represented by arrow $1$.

Given that absorption of photon takes place from an excited state (a rare event at low temperatures).

In the given diagram absorption of photon takes place from the excited state ${\mathrm{E}}_1$to ${\mathrm{E}}_3$.

Therefore the absorption from the excited state ( a rare event at low temperature) is represented by arrow .

Given that the photon is getting emitted with the energy of ${\mathrm{E}}_3−{\mathrm{E}}_1$.

For the emission process the tail of the arrow should be in the initial state and the head of the arrow should be in the final state. As here energy is ${\mathrm{E}}_3−{\mathrm{E}}_1$, so the tail of the arrow should be at ${\mathrm{E}}_3$and the head of the arrow should be at ${\mathrm{E}}_1$.

Therefore, emission of photon whose energy is ${\mathrm{E}}_3−{\mathrm{E}}_1$is represented by arrow $2$.

Given that the photon is getting emitted with the energy of ${\mathrm{E}}_2−{\mathrm{E}}_0$.

For the emission process the tail of the arrow should be in the initial state and the head of the arrow should be in the final state. As here energy is ${\mathrm{E}}_2−{\mathrm{E}}_0$, so the tail of the arrow should be at ${\mathrm{E}}_2$ and the head of the arrow should be at ${\mathrm{E}}_0$.

Therefore emission of photon whose energy is ${\mathrm{E}}_2−{\mathrm{E}}_0$ is represented by arrow $3$.

1. In case of the absorption, the head of the arrow points up, whereas in case of emission the head of the arrow pints down. So the given statement is false.
2. In case of emission the head of the arrow pints down. So the given statement is true.
3. In case of the absorption, the head of the arrow points up. So the given statement is true.
4. For the emission process the tail of the arrow should be in the initial state. So the given statement is true.
5. For the emission or absorption, the head of the drawn to the final state. So the given statement is true.
6. For the emission or absorption, the head of the drawn to the final state. If there is no error, so we can’t be able to find whether the transition is emission or absorption. So the given statement is false.