Calculate the Wavelength of the Spectra when the electron in the atom gains a photon and makes a transition from the ground -state to the excited state?

The wavelength of the spectra can be calculated using the formula:

λ = hc / E

Where:
λ = wavelength of the spectra
h = Planck's constant (6.626 x 10^-34 J s)
c = speed of light (3.00 x 10^8 m/s)
E = energy of the photon gained by the electron during the transition

The energy of the photon gained by the electron during the transition can be calculated using the formula:

E = E2 - E1

Where:
E2 = energy of the excited state
E1 = energy of the ground state

Once the energy of the photon is calculated, it can be substituted into the first formula to find the wavelength of the spectra.

The wavelength of the absorbed photon is given by:

```
λ = hc / ΔE
```

where:

- `λ` is the wavelength of the photon
- `h` is Planck's constant
- `c` is the speed of light
- `ΔE` is the energy difference between the excited state and the ground state

ΔE can be calculated using the following formula:

```
ΔE = E_excited - E_ground
```

where:

- `E_excited` is the energy of the excited state
- `E_ground` is the energy of the ground state

For an electron in an atom, the energy of the ground state is:

```
E_ground = -13.6 eV
```

And the energy of the excited state is:

```
E_excited = -3.4 eV
```

Therefore, the ΔE is:

```
ΔE = (-3.4 eV) - (-13.6 eV) = 10.2 eV
```

Converting eV to J:

```
ΔE = 10.2 eV - 1.602 - 10^-19 J/eV = 1.63 - 10^-18 J
```

Now we can calculate the wavelength of the absorbed photon:

```
λ = hc / ΔE = (6.626 - 10^-34 J s) - (3 - 10^8 m/s) / (1.63 - 10^-18 J) = 1.21 - 10^-7 m
```

Therefore, the wavelength of the spectra when the electron in the atom gains a photon and makes a transition from the ground state to the excited state is 1.21 - 10^-7 m.