If the kinetic energy of an electron is increased 4 times, the wavelength of the de Broglie wave associated with it would becomes:

Options

A. 4 times
B. 2 times
C. `(1)/(2)` times
D. `(1)/(4)` times

Explanation

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

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

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

### Step 1: Write the initial expression for de Broglie wavelength

The initial wavelength (\(\lambda_1\)) when the kinetic energy is \(E\) is:

\[ \lambda_1 = \frac{h}{\sqrt{2 m E}} \]

### Step 2: Consider the change in kinetic energy

According to the problem, the kinetic energy of the electron is increased 4 times. Therefore, the new kinetic energy (\(E'\)) becomes:

\[ E' = 4E \]

### Step 3: Write the expression for the new de Broglie wavelength

Now, we can find the new wavelength (\(\lambda_2\)) using the new kinetic energy:

\[ \lambda_2 = \frac{h}{\sqrt{2 m E'}} \]

Substituting \(E' = 4E\) into the equation gives:

\[ \lambda_2 = \frac{h}{\sqrt{2 m (4E)}} \]

### Step 4: Simplify the expression

This simplifies to:

\[ \lambda_2 = \frac{h}{\sqrt{8 m E}} = \frac{h}{2\sqrt{2 m E}} \]

### Step 5: Relate the new wavelength to the initial wavelength

Now, we can relate \(\lambda_2\) to \(\lambda_1\):

\[ \lambda_2 = \frac{1}{2} \cdot \frac{h}{\sqrt{2 m E}} = \frac{1}{2} \lambda_1 \]

### Conclusion

Thus, if the kinetic energy of the electron is increased 4 times, the wavelength of the de Broglie wave associated with it becomes half of the original wavelength:

\[ \lambda_2 = \frac{1}{2} \lambda_1 \]

### Final Answer

The wavelength of the de Broglie wave associated with the electron would become half of its original wavelength.

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