He atom can be excited to 1s¹ 2p¹ by wavelength 58.44 nm. If lowest excited state for He lies 4857cm^-1 below the above. Calculate the energy for the lower excitation state.

To find the energy for the lower excitation state of helium (He), we need to first understand the relationship between wavelength, energy, and the given information about the excited states. The problem states that helium can be excited to the state 1s² 2p¹ by a wavelength of 58.44 nm, and the lowest excited state lies 4857 cm⁻¹ below this state. Let's break this down step by step.

Step 1: Calculate the Energy of the Excited State

The energy of a photon can be calculated using the formula:

E = hc/λ

Where:

• E = energy of the photon (in joules)
• h = Planck's constant (6.626 x 10⁻³⁴ J·s)
• c = speed of light (3.00 x 10⁸ m/s)
• λ = wavelength (in meters)

First, we need to convert the wavelength from nanometers to meters:

58.44 nm = 58.44 x 10⁻⁹ m

Now, substituting the values into the energy formula:

E = (6.626 x 10⁻³⁴ J·s) * (3.00 x 10⁸ m/s) / (58.44 x 10⁻⁹ m)

Calculating this gives:

E ≈ 3.40 x 10⁻¹⁹ J

Step 2: Convert Energy to Wavenumbers

Next, we need to convert this energy into wavenumbers (cm⁻¹), since the problem states that the lower excited state is 4857 cm⁻¹ below the excited state. The conversion from energy in joules to wavenumbers is given by:

Energy (cm⁻¹) = E (J) * (1 / hc) * (100)

Where:

• 1/hc = 1.986 x 10⁻²³ J·cm

Now, substituting the energy we calculated:

Energy (cm⁻¹) = (3.40 x 10⁻¹⁹ J) * (1 / (1.986 x 10⁻²³ J·cm))

Calculating this gives:

Energy (cm⁻¹) ≈ 17161.5 cm⁻¹

Step 3: Calculate the Energy of the Lower Excitation State

Now that we have the energy of the excited state in wavenumbers, we can find the energy of the lower excitation state. Since the lowest excited state is 4857 cm⁻¹ below the excited state, we subtract this value:

Energy of lower state = 17161.5 cm⁻¹ - 4857 cm⁻¹

Calculating this gives:

Energy of lower state ≈ 12304.5 cm⁻¹

Final Result

Thus, the energy for the lower excitation state of helium is approximately 12304.5 cm⁻¹. This value represents the energy difference between the two states and is crucial for understanding the electronic transitions in helium.