The ratio of `E _(2) - E_(1)` to` E_(4) - E_(3)` for the hydrogen atom is approximately equal to -

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 hydrogen atom, we can use the formula for the energy levels of a hydrogen atom:

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

where \( E_n \) is the energy of the electron in the nth shell and \( n \) is the principal quantum number.

### Step 1: Calculate \( E_2 \) and \( E_1 \)

Using the formula:

- For \( n = 2 \):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

- For \( n = 1 \):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

### Step 2: Calculate \( E_2 - E_1 \)

Now, we can find \( E_2 - E_1 \):
\[ E_2 - E_1 = -3.4 \, \text{eV} - (-13.6 \, \text{eV}) = -3.4 + 13.6 = 10.2 \, \text{eV} \]

### Step 3: Calculate \( E_4 \) and \( E_3 \)

Using the same formula:

- For \( n = 4 \):
\[ E_4 = -\frac{13.6 \, \text{eV}}{4^2} = -\frac{13.6 \, \text{eV}}{16} = -0.85 \, \text{eV} \]

- For \( n = 3 \):
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \]

### Step 4: Calculate \( E_4 - E_3 \)

Now, we can find \( E_4 - E_3 \):
\[ E_4 - E_3 = -0.85 \, \text{eV} - (-1.51 \, \text{eV}) = -0.85 + 1.51 = 0.66 \, \text{eV} \]

### Step 5: Calculate the ratio \( \frac{E_2 - E_1}{E_4 - E_3} \)

Now we can find the ratio:
\[ \frac{E_2 - E_1}{E_4 - E_3} = \frac{10.2 \, \text{eV}}{0.66 \, \text{eV}} \approx 15.45 \]

### Step 6: Approximate the result

The approximate value of this ratio is:
\[ \approx 15 \]

Thus, the ratio of \( E_2 - E_1 \) to \( E_4 - E_3 \) for the hydrogen atom is approximately equal to **15**.

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