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The energy of an electron in the first ...

The energy of an electron in the first level of H atom is `- 13.6 eV` .The possible values of the excited states for electron in `He^(o+)` is (are) :

A

`-54.4 eV`

B

`-13.6 eV`

C

`-3.4 eV`

D

`-6.4 eV`

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To determine the possible values of the excited states for the electron in the He^(o+) ion, we can use the formula for the energy of an electron in a hydrogen-like atom: \[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \] Where: - \( E_n \) is the energy of the electron at the principal quantum number \( n \). - \( Z \) is the atomic number of the element (for He, \( Z = 2 \)). - \( n \) is the principal quantum number (1, 2, 3,...). ### Step 1: Calculate the energy for the first level (n=1) For the first level (n=1): \[ E_1 = -\frac{13.6 \cdot 2^2}{1^2} \] \[ E_1 = -\frac{13.6 \cdot 4}{1} \] \[ E_1 = -54.4 \, \text{eV} \] ### Step 2: Calculate the energy for the second level (n=2) For the second level (n=2): \[ E_2 = -\frac{13.6 \cdot 2^2}{2^2} \] \[ E_2 = -\frac{13.6 \cdot 4}{4} \] \[ E_2 = -13.6 \, \text{eV} \] ### Step 3: Calculate the energy for the third level (n=3) For the third level (n=3): \[ E_3 = -\frac{13.6 \cdot 2^2}{3^2} \] \[ E_3 = -\frac{13.6 \cdot 4}{9} \] \[ E_3 = -\frac{54.4}{9} \] \[ E_3 \approx -6.04 \, \text{eV} \] ### Step 4: Calculate the energy for the fourth level (n=4) For the fourth level (n=4): \[ E_4 = -\frac{13.6 \cdot 2^2}{4^2} \] \[ E_4 = -\frac{13.6 \cdot 4}{16} \] \[ E_4 = -\frac{54.4}{16} \] \[ E_4 = -3.4 \, \text{eV} \] ### Summary of Results - \( E_1 = -54.4 \, \text{eV} \) (not an excited state) - \( E_2 = -13.6 \, \text{eV} \) (not an excited state) - \( E_3 \approx -6.04 \, \text{eV} \) (excited state) - \( E_4 = -3.4 \, \text{eV} \) (excited state) ### Conclusion The possible values of the excited states for the electron in He^(o+) are: - \( E_3 \approx -6.04 \, \text{eV} \) - \( E_4 = -3.4 \, \text{eV} \) ### Final Answer The correct options for the excited states are: - Option 3: \( -6.04 \, \text{eV} \) - Option 4: \( -3.4 \, \text{eV} \)

To determine the possible values of the excited states for the electron in the He^(o+) ion, we can use the formula for the energy of an electron in a hydrogen-like atom: \[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \] Where: - \( E_n \) is the energy of the electron at the principal quantum number \( n \). - \( Z \) is the atomic number of the element (for He, \( Z = 2 \)). - \( n \) is the principal quantum number (1, 2, 3,...). ...
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