He atom can be excited to `1s^(1) 2p^(1)` by `lambda=58.44nm`. If lowest excited state for He lies `4857cm^(-1)` below the above. Calculate the energy for the lower excitation state.

Answer

To solve the problem step by step, we will calculate the energy of the excited state of the helium atom and then find the energy for the lower excitation state.

### Step 1: Calculate the energy of the excited state using the wavelength
We use the formula for energy:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 4.135667696 \times 10^{-15} \) eV·s),
- \( c \) is the speed of light (\( 3 \times 10^8 \) m/s),
- \( \lambda \) is the wavelength in meters.

Given:
- \( \lambda = 58.44 \) nm = \( 58.44 \times 10^{-9} \) m.

Now, substituting the values:
\[ E = \frac{(4.135667696 \times 10^{-15} \text{ eV·s})(3 \times 10^8 \text{ m/s})}{58.44 \times 10^{-9} \text{ m}} \]

Calculating this gives:
\[ E \approx 21.25 \text{ eV} \]

### Step 2: Convert the energy difference from cm\(^{-1}\) to eV
The problem states that the lowest excited state lies \( 4857 \) cm\(^{-1} \) below the excited state. We need to convert this to eV.

Using the conversion:
\[ 1 \text{ cm}^{-1} = 1.23984 \times 10^{-4} \text{ eV} \]
Thus,
\[ E_{\text{difference}} = 4857 \text{ cm}^{-1} \times 1.23984 \times 10^{-4} \text{ eV/cm}^{-1} \]

Calculating this gives:
\[ E_{\text{difference}} \approx 0.602 \text{ eV} \]

### Step 3: Calculate the energy of the lower excitation state
The energy of the lower excitation state can be calculated by subtracting the energy difference from the energy of the higher excited state:
\[ E_{\text{lower}} = E_{\text{higher}} - E_{\text{difference}} \]
\[ E_{\text{lower}} = 21.25 \text{ eV} - 0.602 \text{ eV} \]

Calculating this gives:
\[ E_{\text{lower}} \approx 20.648 \text{ eV} \]

### Final Answer
The energy for the lower excitation state is approximately **20.648 eV**.

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