When a charged particle is accelerated through 1.5volts its de -Broglie wavelength is `lamda_(1)` and it is `lamda_(2)` when the same particle is accelerated through 13.5 volts. The ratio `lamda_(1)//lamda_(2)`is……………

Explanation

To solve the problem, we need to find the ratio of the de Broglie wavelengths \( \lambda_1 \) and \( \lambda_2 \) when a charged particle is accelerated through different voltages.

### Step 1: Understand the relationship between de Broglie wavelength and potential
The de Broglie wavelength \( \lambda \) of a particle can be expressed as:
\[ \lambda = \frac{h}{\sqrt{2m eV}} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( e \) is the charge of the particle,
- \( V \) is the accelerating voltage.

### Step 2: Establish the relationship between wavelengths and voltages
From the formula, we can see that the de Broglie wavelength is inversely proportional to the square root of the voltage:
\[ \lambda \propto \frac{1}{\sqrt{V}} \]
Thus, we can write the ratio of the wavelengths as:
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}} \]

### Step 3: Substitute the given voltages
Given:
- \( V_1 = 1.5 \) volts
- \( V_2 = 13.5 \) volts

Now substituting the values into the equation:
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{13.5}{1.5}} \]

### Step 4: Simplify the ratio
Calculating the fraction:
\[ \frac{13.5}{1.5} = 9 \]
Thus, we have:
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{9} = 3 \]

### Step 5: Final answer
The ratio \( \frac{\lambda_1}{\lambda_2} \) is:
\[ \frac{\lambda_1}{\lambda_2} = 3 \]

### Conclusion
The final result is:
\[ \lambda_1 : \lambda_2 = 3 : 1 \]