Satement-1: Emitted radiation will fall in visible range when an electron jumps from higher level ot `n=2 "in"Li^(+2) "ion".`
Statement-2: Balmer series radiations belong to visible range in all H-atoms.

Answer

To solve the question, we need to analyze both statements provided about the transitions of electrons in the lithium ion (Li²⁺) and the hydrogen atom.

### Step-by-Step Solution:

1. **Understanding Statement 1**:
- Statement 1 claims that emitted radiation will fall in the visible range when an electron jumps from a higher energy level to \( n=2 \) in the lithium ion \( Li^{+2} \).
- For an electron to emit radiation in the visible range, the transition must correspond to energy levels that produce wavelengths within the visible spectrum (approximately 400 nm to 700 nm).

2. **Analyzing the Energy Levels of \( Li^{+2} \)**:
- The energy levels of hydrogen-like ions can be calculated using the formula:
\[ E_n = -\frac{Z^2 \cdot R_H}{n^2} \]
where \( Z \) is the atomic number, \( R_H \) is the Rydberg constant, and \( n \) is the principal quantum number.
- For \( Li^{+2} \), \( Z = 3 \). Thus, the energy levels will be:
\[ E_n = -\frac{3^2 \cdot R_H}{n^2} = -\frac{9 \cdot R_H}{n^2} \]

3. **Calculating the Wavelength for Transitions to \( n=2 \)**:
- The transition from a higher level \( n \) to \( n=2 \) will have a wavelength given by the Rydberg formula:
\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \]
- For \( Li^{+2} \):
\[ \frac{1}{\lambda} = R_H \cdot 9 \left( \frac{1}{4} - \frac{1}{n^2} \right) \]

4. **Determining the Visible Range**:
- As \( n \) increases, \( \frac{1}{n^2} \) decreases, which means the term \( \left( \frac{1}{4} - \frac{1}{n^2} \right) \) becomes larger, leading to shorter wavelengths (higher energy).
- The energy of emitted radiation will be significantly higher than that of the hydrogen atom due to the factor of \( Z^2 \) (which is 9 for lithium). This means that the emitted radiation for transitions to \( n=2 \) will not fall in the visible range, but rather in the ultraviolet range.

5. **Conclusion for Statement 1**:
- Therefore, Statement 1 is **false** because the emitted radiation does not fall in the visible range for \( Li^{+2} \).

6. **Understanding Statement 2**:
- Statement 2 claims that Balmer series radiations belong to the visible range in all hydrogen atoms.
- The Balmer series corresponds to transitions from higher energy levels to \( n=2 \) in hydrogen, which indeed produces visible light (specifically, wavelengths in the range of 400 nm to 700 nm).

7. **Conclusion for Statement 2**:
- Thus, Statement 2 is **true**.

### Final Answer:
- Statement 1 is false, and Statement 2 is true.