He atom can be excited to 1s1 2p1 by Lambda = 58.44 nm. If the lowest excited state for He lies 4857 cm ^-1 below the above, calculate te enrgy for the lower excitation state.

To find the energy for the lower excitation state of helium (He) after it has been excited to the state 1s1 2p1, we first need to understand the relationship between wavelength, energy, and the energy levels of the atom. The energy of a photon can be calculated using the formula:

Energy Calculation from Wavelength

The energy (E) of a photon is given by the equation:

E = hc / λ

Where:

• h is Planck's constant (6.626 x 10^-34 J·s)
• c is the speed of light (3.00 x 10^8 m/s)
• λ is the wavelength in meters

Given that the wavelength (λ) is 58.44 nm, we first convert this to meters:

58.44 nm = 58.44 x 10^-9 m

Calculating the Energy for 1s1 2p1 State

Now, substituting the values into the energy equation:

E = (6.626 x 10^-34 J·s) * (3.00 x 10^8 m/s) / (58.44 x 10^-9 m)

Calculating this gives:

E ≈ 3.40 x 10^-14 J

Finding the Energy of the Lower Excited State

Next, we need to account for the energy difference between the excited state (1s1 2p1) and the lower excited state. The problem states that the lowest excited state for He lies 4857 cm^-1 below the 1s1 2p1 state. To convert this wavenumber to energy, we use the formula:

E = hc * ν

Where ν (nu) is the wavenumber in reciprocal centimeters (cm^-1). First, we convert the wavenumber to meters:

ν = 4857 cm^-1 = 4857 x 100 = 485700 m^-1

Calculating the Energy Difference

Now we can calculate the energy difference:

E = (6.626 x 10^-34 J·s) * (3.00 x 10^8 m/s) * (485700 m^-1)

Calculating this gives:

E ≈ 9.67 x 10^-19 J

Final Energy for the Lower Excited State

To find the energy of the lower excitation state, we subtract the energy difference from the energy of the 1s1 2p1 state:

E_lower = E_1s1 2p1 - E_difference

Substituting the values:

E_lower = (3.40 x 10^-14 J) - (9.67 x 10^-19 J)

Since the energy difference is much smaller than the energy of the excited state, we can approximate:

E_lower ≈ 3.40 x 10^-14 J

This value represents the energy of the lower excited state of helium after accounting for the energy difference due to the transition. In summary, the energy for the lower excitation state of helium is approximately 3.40 x 10^-14 J, considering the specified energy difference.