The ionisation enthalpy of hydrogen atom is `1.312 xx 10^6" J mol"^(-1)`. The energy required to excite the electron in the atom from n = 1 to n = 2 is

Options

A. `8.51xx10^(5)"J mol"^(-1)`
B. `6.56 xx10^(5)"J mol"^(-1)`
C. `7.56 xx10^(5)"J mol"^(-1)`
D. `9.84 xx10^(5)"J mol"^(-1)`

Explanation

To solve the problem of finding the energy required to excite the electron in a hydrogen atom from n = 1 to n = 2, we can follow these steps:

### Step 1: Understand the Ionization Enthalpy
The ionization enthalpy of hydrogen is given as \(1.312 \times 10^6 \, \text{J mol}^{-1}\). This value represents the energy required to remove the electron completely from the ground state (n=1) to infinity (n=∞).

### Step 2: Use the Formula for Energy Levels
The energy of an electron in a hydrogen atom at a given principal quantum number \(n\) is given by the formula:
\[ E_n = -\frac{R_H}{n^2} \]
where \(R_H\) is the Rydberg constant for hydrogen, which can be calculated from the ionization enthalpy:
\[ R_H = 1.312 \times 10^6 \, \text{J mol}^{-1} \]

### Step 3: Calculate the Energy for n = 1 and n = 2
Now, we will calculate the energies for \(n = 1\) and \(n = 2\):
- For \(n = 1\):
\[ E_1 = -\frac{1.312 \times 10^6}{1^2} = -1.312 \times 10^6 \, \text{J mol}^{-1} \]
- For \(n = 2\):
\[ E_2 = -\frac{1.312 \times 10^6}{2^2} = -\frac{1.312 \times 10^6}{4} = -3.28 \times 10^5 \, \text{J mol}^{-1} \]

### Step 4: Calculate the Energy Change
The energy required to excite the electron from \(n = 1\) to \(n = 2\) is given by the difference in energy levels:
\[ \Delta E = E_2 - E_1 \]
Substituting the values we calculated:
\[ \Delta E = (-3.28 \times 10^5) - (-1.312 \times 10^6) \]
\[ \Delta E = -3.28 \times 10^5 + 1.312 \times 10^6 \]
\[ \Delta E = 1.312 \times 10^6 - 3.28 \times 10^5 = 9.84 \times 10^5 \, \text{J mol}^{-1} \]

### Step 5: Conclusion
Thus, the energy required to excite the electron in the hydrogen atom from \(n = 1\) to \(n = 2\) is:
\[ \Delta E = 9.84 \times 10^5 \, \text{J mol}^{-1} \]

### Final Answer
The correct option is:
**9.84 × 10^5 J mol^(-1)**

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