To which quantum level does the electron jump in H atom from the lowest level if it is given an energy corresponding to `99%` of the ionization potential of hydrogen atom?

Explanation

To solve the problem of determining to which quantum level an electron in a hydrogen atom jumps when given energy corresponding to 99% of the ionization potential, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Ionization Energy**: The ionization potential (or ionization energy) of the hydrogen atom is the energy required to remove the electron from the ground state (n=1) to infinity (n=∞). The ionization energy of hydrogen is approximately 13.6 eV.

2. **Calculate 99% of Ionization Energy**:
\[ \text{Energy given} = 0.99 \times \text{Ionization Energy} = 0.99 \times 13.6 \, \text{eV} = 13.464 \, \text{eV} \]

3. **Use the Energy Level Formula**: The energy of an electron in a hydrogen atom at quantum level \( n \) is given by:
\[ E_n = -\frac{E_0}{n^2} \]
where \( E_0 = 13.6 \, \text{eV} \).

4. **Set Up the Equation**: The energy difference when the electron jumps from level \( n_1 \) (where \( n_1 = 1 \)) to level \( n_2 \) can be expressed as:
\[ E = E_{n_2} - E_{n_1} = -\frac{E_0}{n_2^2} - \left(-\frac{E_0}{1^2}\right) = E_0 \left(1 - \frac{1}{n_2^2}\right) \]

5. **Substitute the Given Energy**: We know that the energy given is 99% of the ionization energy, so we set:
\[ 13.464 = 13.6 \left(1 - \frac{1}{n_2^2}\right) \]

6. **Solve for \( n_2 \)**:
- Divide both sides by 13.6:
\[ \frac{13.464}{13.6} = 1 - \frac{1}{n_2^2} \]
- Calculate \( \frac{13.464}{13.6} \):
\[ 0.99 = 1 - \frac{1}{n_2^2} \]
- Rearranging gives:
\[ \frac{1}{n_2^2} = 1 - 0.99 = 0.01 \]
- Taking the reciprocal gives:
\[ n_2^2 = 100 \]
- Taking the square root gives:
\[ n_2 = 10 \]

### Final Answer:
The electron jumps to the quantum level \( n = 10 \).