In hydrogen atom, if the difference in the energy of the electron in n =2 and n=3 orbits is E, the ionization energy of hydrogen atom is

Options

A. 13.2E
B. 7.2E
C. 5.6E
D. 3.2E

Answer

To find the ionization energy of a hydrogen atom given the difference in energy between the n=2 and n=3 orbits as E, we can follow these steps:

### Step 1: Understand the Energy Levels
The energy of an electron in a hydrogen atom at a given orbit (n) is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

### Step 2: Calculate the Energy for n=2 and n=3
Using the formula, we can calculate the energy for the n=2 and n=3 levels.

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

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

### Step 3: Find the Energy Difference
The difference in energy between the n=2 and n=3 orbits is:
\[ E = E_3 - E_2 = (-1.51 \, \text{eV}) - (-3.4 \, \text{eV}) = -1.51 + 3.4 = 1.89 \, \text{eV} \]

### Step 4: Relate the Energy Difference to Ionization Energy
The ionization energy (I) of the hydrogen atom is the energy required to remove the electron completely from the ground state (n=1). The formula for the ionization energy can be expressed as:
\[ I = -E_1 = -\left(-\frac{13.6 \, \text{eV}}{1^2}\right) = 13.6 \, \text{eV} \]

### Step 5: Express Ionization Energy in Terms of E
From the energy difference calculated earlier, we can express the ionization energy in terms of E. The relationship derived from the energy levels is:
\[ E = I \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating the right-hand side:
\[ \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus, we have:
\[ E = I \cdot \frac{5}{36} \]
Rearranging gives:
\[ I = E \cdot \frac{36}{5} \]

### Step 6: Substitute E
Substituting the value of E, we find:
\[ I = \frac{36}{5} E \approx 7.2 E \]

### Final Answer
Thus, the ionization energy of the hydrogen atom is:
\[ \boxed{7.2E} \]