A proton and a `Li^(3+)` nucleus are accelerated by the same potential. If `lambda_(Li)` and `lambda_(rho)` denote the de Broglie wavelengths of Lit and proton respectively, then the value `lambda_(Li)/lambda_rho" is ""x"xx10^(-1)`. The value of x is `"______"`.

Answer

To solve the problem, we need to find the ratio of the de Broglie wavelengths of a proton and a lithium-3 ion (Li³⁺) that have been accelerated through the same potential.

### Step-by-Step Solution:

1. **Understand the de Broglie wavelength formula**:
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2mqV}} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( q \) is the charge of the particle,
- \( V \) is the potential through which the particle is accelerated.

2. **Write the expressions for the wavelengths**:
For the lithium ion (Li³⁺), we have:
\[ \lambda_{Li} = \frac{h}{\sqrt{2m_{Li}q_{Li}V}} \]
For the proton (p), we have:
\[ \lambda_{p} = \frac{h}{\sqrt{2m_{p}q_{p}V}} \]

3. **Identify the masses and charges**:
- The mass of a proton (\( m_p \)) is approximately \( 1 \, \text{u} \) (atomic mass unit).
- The mass of lithium-3 ion (\( m_{Li} \)) is approximately \( 3 \, \text{u} \) (since it has 3 nucleons).
- The charge of a proton (\( q_p \)) is \( +1e \).
- The charge of lithium-3 ion (\( q_{Li} \)) is \( +3e \).

4. **Substitute the values into the wavelength formulas**:
For lithium:
\[ \lambda_{Li} = \frac{h}{\sqrt{2 \cdot 3 \, \text{u} \cdot 3e \cdot V}} \]
For the proton:
\[ \lambda_{p} = \frac{h}{\sqrt{2 \cdot 1 \, \text{u} \cdot 1e \cdot V}} \]

5. **Calculate the ratio of the wavelengths**:
\[ \frac{\lambda_{Li}}{\lambda_{p}} = \frac{\frac{h}{\sqrt{2 \cdot 3 \, \text{u} \cdot 3e \cdot V}}}{\frac{h}{\sqrt{2 \cdot 1 \, \text{u} \cdot 1e \cdot V}}} \]
Simplifying this gives:
\[ \frac{\lambda_{Li}}{\lambda_{p}} = \frac{\sqrt{2 \cdot 1 \, \text{u} \cdot 1e \cdot V}}{\sqrt{2 \cdot 3 \, \text{u} \cdot 3e \cdot V}} = \frac{\sqrt{1}}{\sqrt{9}} = \frac{1}{3} \]

6. **Express the result in the required form**:
We need to express this in the form \( x \times 10^{-1} \):
\[ \frac{\lambda_{Li}}{\lambda_{p}} = \frac{1}{3} = \frac{0.1}{3} \approx 0.333 \approx 3.33 \times 10^{-1} \]
Thus, \( x \) is approximately \( 3.33 \), but we round it to \( 3 \).

### Final Answer:
The value of \( x \) is \( 3 \).