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

Answer

To solve the problem, we need to calculate the energy for the lower excitation state of the helium atom after it is excited to the state \(1s^1 2p^1\) using the given wavelength and the energy difference provided.

### Step-by-Step Solution:

1. **Calculate the Energy of the Excited State:**
The energy \(E\) corresponding to a wavelength \(\lambda\) can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h\) (Planck's constant) = \(6.626 \times 10^{-34} \, \text{J s}\)
- \(c\) (speed of light) = \(3 \times 10^8 \, \text{m/s}\)
- \(\lambda = 58.44 \, \text{nm} = 58.44 \times 10^{-9} \, \text{m}\)

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{58.44 \times 10^{-9} \, \text{m}} \]
\[ E \approx 3.41 \times 10^{-19} \, \text{J} \]

To convert this energy into electron volts (1 eV = \(1.602 \times 10^{-19} \, \text{J}\)):
\[ E \approx \frac{3.41 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \approx 21.25 \, \text{eV} \]

2. **Calculate the Energy of the Lowest Excited State:**
The lowest excited state for helium is given as \(4857 \, \text{cm}^{-1}\). We need to convert this to energy in electron volts. The energy in terms of wavenumber (\( \tilde{\nu} \)) is given by:
\[ E = \tilde{\nu} \cdot hc \]
where \( \tilde{\nu} = 4857 \, \text{cm}^{-1} = 4857 \times 100 \, \text{m}^{-1} = 485700 \, \text{m}^{-1} \).

Substituting the values:
\[ E = (485700 \, \text{m}^{-1})(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s}) \]
\[ E \approx 9.66 \times 10^{-19} \, \text{J} \]

Converting this to electron volts:
\[ E \approx \frac{9.66 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \approx 6.03 \, \text{eV} \]

3. **Calculate the Energy of the Lower Excitation State:**
The energy of the lower excitation state is the energy of the excited state minus the energy of the lowest excited state:
\[ E_{\text{lower}} = E_{\text{excited}} - E_{\text{lowest}} \]
\[ E_{\text{lower}} = 21.25 \, \text{eV} - 6.03 \, \text{eV} \approx 15.22 \, \text{eV} \]

### Final Answer:
The energy for the lower excitation state of the helium atom is approximately **15.22 eV**.

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