The ratio of `(E_(2) - E_(1))` to `(E_(4) - E_(3))` for `He^(+)` ion is approximately equal to (where `E_(n)` is the energy of nth orbit ):

Options

A. 10
B. 15
C. 17
D. 12

Answer

To find the ratio of \((E_{2} - E_{1})\) to \((E_{4} - E_{3})\) for the \(He^{+}\) ion, we will use the formula for the energy levels of hydrogen-like atoms:

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

where \(Z\) is the atomic number and \(n\) is the principal quantum number.

### Step 1: Calculate \(E_{2}\) and \(E_{1}\)

For \(He^{+}\), the atomic number \(Z = 2\).

- Calculate \(E_{2}\):
\[ E_{2} = -\frac{13.6 \times 2^2}{2^2} = -\frac{13.6 \times 4}{4} = -13.6 \, \text{eV} \]

- Calculate \(E_{1}\):
\[ E_{1} = -\frac{13.6 \times 2^2}{1^2} = -\frac{13.6 \times 4}{1} = -54.4 \, \text{eV} \]

Now, calculate \(E_{2} - E_{1}\):
\[ E_{2} - E_{1} = -13.6 - (-54.4) = -13.6 + 54.4 = 40.8 \, \text{eV} \]

### Step 2: Calculate \(E_{4}\) and \(E_{3}\)

- Calculate \(E_{4}\):
\[ E_{4} = -\frac{13.6 \times 2^2}{4^2} = -\frac{13.6 \times 4}{16} = -3.4 \, \text{eV} \]

- Calculate \(E_{3}\):
\[ E_{3} = -\frac{13.6 \times 2^2}{3^2} = -\frac{13.6 \times 4}{9} = -6.04 \, \text{eV} \]

Now, calculate \(E_{4} - E_{3}\):
\[ E_{4} - E_{3} = -3.4 - (-6.04) = -3.4 + 6.04 = 2.64 \, \text{eV} \]

### Step 3: Calculate the ratio \(\frac{(E_{2} - E_{1})}{(E_{4} - E_{3})}\)

Now we can find the ratio:
\[ \frac{(E_{2} - E_{1})}{(E_{4} - E_{3})} = \frac{40.8}{2.64} \]

Calculating the ratio:
\[ \frac{40.8}{2.64} \approx 15.0 \]

### Final Answer
The ratio of \((E_{2} - E_{1})\) to \((E_{4} - E_{3})\) for the \(He^{+}\) ion is approximately equal to **15**.

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