The difference in energy between the 2nd & 3rd orbit of `He^+` ion will be

Answer

To find the difference in energy between the 2nd and 3rd orbits of the \( He^+ \) ion, we can use the formula for the energy levels of hydrogen-like ions:

\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]

where:
- \( E_n \) is the energy of the nth orbit,
- \( Z \) is the atomic number,
- \( n \) is the principal quantum number.

For the \( He^+ \) ion, the atomic number \( Z \) is 2.

### Step 1: Calculate the energy of the 2nd orbit (\( E_2 \))

Using the formula for \( n = 2 \):

\[ E_2 = -\frac{13.6 \, \text{eV} \cdot 2^2}{2^2} = -\frac{13.6 \, \text{eV} \cdot 4}{4} = -13.6 \, \text{eV} \]

### Step 2: Calculate the energy of the 3rd orbit (\( E_3 \))

Using the formula for \( n = 3 \):

\[ E_3 = -\frac{13.6 \, \text{eV} \cdot 2^2}{3^2} = -\frac{13.6 \, \text{eV} \cdot 4}{9} = -6.04 \, \text{eV} \]

### Step 3: Calculate the difference in energy between the 3rd and 2nd orbits

The difference in energy (\( \Delta E \)) is given by:

\[ \Delta E = E_3 - E_2 \]

Substituting the values we found:

\[ \Delta E = (-6.04 \, \text{eV}) - (-13.6 \, \text{eV}) = -6.04 + 13.6 = 7.56 \, \text{eV} \]

### Final Answer

The difference in energy between the 2nd and 3rd orbit of the \( He^+ \) ion is \( 7.56 \, \text{eV} \).

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