Question

An electron in the 7th energy level of the H atom drops to the 3th energy level. In other words an electron in an excited state drops to a less excited state. What is the energy (in J) of the emitted photon? The energy of an electron in the nth level is_________.

Answer 1

Step-by-step explanation:

The given data is as follows.

`n_(1)` = 7,
`n_(2)` = 3

Z for H = 1

According to Reydberg's equation, we will calculate the energy emitted by the photon as follows.

`\Delta E = -2.178 * 10^(-18) J * (Z)^(2)[(1)/(n^(2)_(2)) - (1)/(n^(2)_(1))]`

=
`-2.178 * 10^(-18) J * (1)^(2)[(1)/((3)^(2)) - (1)/((7)^(2))]`

=
`-2.178 * 10^(-18) J * (0.11 - 0.02)`

=
`-1.96 * 10^(-19) J`

The negative sign indicates that energy is released in the process.

Thus, we can conclude that energy (in J) of the emitted photon is
`-1.96 * 10^(-19) J`.