Supposing the energy of fourth shell for hydrogen atom is - 50 a.u. (arbitrary unit). What would be its ionization potential -

Options

A. 50
B. 800
C. `15.4`
D. `20.8`

Answer

To find the ionization potential of a hydrogen atom given that the energy of the fourth shell is -50 a.u., we can follow these steps:

### Step 1: Understand the relationship between energy and ionization potential
The energy of an electron in the nth shell of a hydrogen atom is given by the formula:
\[ E_n = -\frac{I}{n^2} \]
where \(E_n\) is the energy of the nth shell, \(I\) is the ionization energy, and \(n\) is the principal quantum number.

### Step 2: Substitute the known values
From the question, we know:
- The energy of the fourth shell (\(E_4\)) is -50 a.u.
- For the fourth shell, \(n = 4\).

Using the formula:
\[ E_4 = -\frac{I}{4^2} \]
Substituting \(E_4 = -50\):
\[ -50 = -\frac{I}{16} \]

### Step 3: Solve for the ionization energy \(I\)
Rearranging the equation to solve for \(I\):
\[ 50 = \frac{I}{16} \]
Multiplying both sides by 16:
\[ I = 50 \times 16 \]
\[ I = 800 \text{ a.u.} \]

### Step 4: Conclusion
The ionization potential of the hydrogen atom is 800 a.u.