Question

Consider these 4 transitions for an electron in a hydrogen atom. I) n = 7 --> n = 1 II) n = 7 --> n = 5 III) n = 1 --> n = 7 IV) n = 7 --> n = 6. Of those that will emit of photon, which will emit a photon with the longest wavelength?

Answer 1

Step-by-step explanation:

Using Reydberg's equation, we will calculate the energy emitted by the photon for the given transitions.

The equation is as follows.

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

1). For
`n_(1)` = 7,
`n_(2)` = 1, Z for H = 1
`\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)/((1)^(2)) - (1)/((7)^(2))]`

=
`-2.178 * 10^(-18) J * (1 - 0.02)`

=
`-12.13 * 10^(-18) J`

As, E =
`(hc)/(\lambda)`

Putting the given values and calculate the wavelength as follows.

E =
`(hc)/(\lambda)`

`12.13 * 10^(-18) J` =
`(6.626 * 10^(-34)Js * 3 * 10^(8 m/s))/(\lambda)`

`\lambda = 1.58 * 10^(-8)` m

=
`15.8 * 10^(-9)` m

= 15.8 nm (as 1 m =
`10^(-9)` nm)

2). For
`n_(1)` = 7,
`n_(2)` = 5, Z for H = 1
`\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)/((5)^(2)) - (1)/((7)^(2))]`

=
`-2.178 * 10^(-18) J * (0.04 - 0.02)`

=
`-0.0435 * 10^(-18) J`

As, E =
`(hc)/(\lambda)`

Putting the given values and calculate the wavelength as follows.

E =
`(hc)/(\lambda)`

`0.0435 * 10^(-18) J` =
`(6.626 * 10^(-34)Js * 3 * 10^(8 m/s))/(\lambda)`

`\lambda = 456.9 * 10^(-8)` m

= 45.6 nm

3). For
`n_(1)` = 1,
`n_(2)` = 7, Z for H = 1
`\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)/((7)^(2)) - (1)/((1)^(2))]`

=
`-2.178 * 10^(-18) J * (0.02 - 1)`

=
`2.13 * 10^(-18) J`

As, E =
`(hc)/(\lambda)`

Putting the given values and calculate the wavelength as follows.

E =
`(hc)/(\lambda)`

`2.13 * 10^(-18) J` =
`(6.626 * 10^(-34)Js * 3 * 10^(8 m/s))/(\lambda)`

`\lambda = 9.33 * 10^(-8)` m

= 93.3 nm

4). For
`n_(1)` = 7,
`n_(2)` = 6, Z for H = 1
`\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)/((6)^(2)) - (1)/((7)^(2))]`

=
`-2.178 * 10^(-18) J * (0.03 - 0.02)`

=
`-0.0217 * 10^(-18) J`

As, E =
`(hc)/(\lambda)`

Putting the given values and calculate the wavelength as follows.

E =
`(hc)/(\lambda)`

`0.0217 * 10^(-18)J` =
`(6.626 * 10^(-34)Js * 3 * 10^(8 m/s))/(\lambda)`

`\lambda = 916.03 * 10^(-8)` m

= 9160.3 nm

Thus, we can conclude that transition from n = 7 to n = 6 will emit a photon with the longest wavelength.