If kinetic energy of an electron is reduced by (1/9) then how many times its de Broglie wavelength will increase.

Answer

To solve the problem, we need to relate the de Broglie wavelength of an electron to its kinetic energy. The de Broglie wavelength (\( \lambda \)) is given by the formula:

\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]

where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the electron,
- \( KE \) is the kinetic energy of the electron.

### Step 1: Understand the relationship between wavelength and kinetic energy
From the formula, we can see that the wavelength is inversely proportional to the square root of the kinetic energy:

\[ \lambda \propto \frac{1}{\sqrt{KE}} \]

### Step 2: Analyze the change in kinetic energy
The problem states that the kinetic energy of the electron is reduced by \( \frac{1}{9} \). This means if the initial kinetic energy is \( KE_1 \), the new kinetic energy \( KE_2 \) can be expressed as:

\[ KE_2 = KE_1 - \frac{1}{9} KE_1 = \frac{8}{9} KE_1 \]

### Step 3: Set up the ratio of wavelengths
Using the relationship established in Step 1, we can express the ratio of the wavelengths before and after the change in kinetic energy:

\[ \frac{\lambda_2}{\lambda_1} = \frac{\sqrt{KE_1}}{\sqrt{KE_2}} \]

### Step 4: Substitute the values of kinetic energy
Substituting \( KE_2 \) from Step 2 into the equation gives:

\[ \frac{\lambda_2}{\lambda_1} = \frac{\sqrt{KE_1}}{\sqrt{\frac{8}{9} KE_1}} = \frac{\sqrt{KE_1}}{\sqrt{KE_1} \cdot \sqrt{\frac{8}{9}}} \]

This simplifies to:

\[ \frac{\lambda_2}{\lambda_1} = \frac{1}{\sqrt{\frac{8}{9}}} = \frac{1}{\frac{2\sqrt{2}}{3}} = \frac{3}{2\sqrt{2}} \]

### Step 5: Calculate the increase in wavelength
To find how many times the wavelength increases, we need to find the reciprocal of the ratio:

\[ \frac{\lambda_1}{\lambda_2} = \frac{2\sqrt{2}}{3} \]

Thus, the increase in wavelength can be expressed as:

\[ \lambda_2 = \frac{3}{2\sqrt{2}} \lambda_1 \]

### Conclusion
The final result shows that the de Broglie wavelength increases by a factor of \( \frac{3}{2\sqrt{2}} \) times the initial wavelength.