An electron is continuously accelerated in a vacuum tube by applying potential differece. If the de-Broglie's wavelength is decreased by `10%`, the change in the kinetic energy of the electron is nearly

Options

A. decreased by `11%`
B. increased by `23.4%`
C. increased by `10%`
D. increased by `11.1%`

Explanation

To solve the problem step by step, we need to analyze the relationship between the de Broglie wavelength and the kinetic energy of the electron.

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

### Step 2: Relate momentum to kinetic energy
The momentum \( p \) of an electron can be expressed in terms of its kinetic energy (KE):
\[ p = \sqrt{2m \cdot KE} \]
where \( m \) is the mass of the electron.

### Step 3: Substitute momentum into the wavelength formula
Substituting the expression for momentum into the de Broglie wavelength formula, we have:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]

### Step 4: Analyze the change in wavelength
If the de Broglie wavelength decreases by 10%, the new wavelength (λ') can be expressed as:
\[ \lambda' = \lambda - 0.1\lambda = 0.9\lambda \]

### Step 5: Set up the equation for the new wavelength
Using the new wavelength in the de Broglie formula:
\[ \lambda' = \frac{h}{\sqrt{2m \cdot KE'}} \]
where \( KE' \) is the new kinetic energy after the change.

### Step 6: Set up the ratio of the wavelengths
Setting the original and new wavelengths equal gives:
\[ 0.9\lambda = \frac{h}{\sqrt{2m \cdot KE'}} \]
Now we can set up the ratio:
\[ \frac{\lambda}{0.9\lambda} = \frac{\sqrt{2m \cdot KE'}}{h} \]
This simplifies to:
\[ \frac{1}{0.9} = \frac{\sqrt{2m \cdot KE'}}{\sqrt{2m \cdot KE}} \]

### Step 7: Square both sides
Squaring both sides gives:
\[ \left(\frac{1}{0.9}\right)^2 = \frac{KE'}{KE} \]
This simplifies to:
\[ \frac{1}{0.81} = \frac{KE'}{KE} \]

### Step 8: Solve for the new kinetic energy
Rearranging gives:
\[ KE' = \frac{KE}{0.81} \approx 1.2346 \cdot KE \]

### Step 9: Calculate the change in kinetic energy
The change in kinetic energy is:
\[ \Delta KE = KE' - KE = 1.2346 \cdot KE - KE = 0.2346 \cdot KE \]

### Step 10: Calculate the percentage change
To find the percentage change in kinetic energy:
\[ \text{Percentage Change} = \left(\frac{\Delta KE}{KE}\right) \times 100 = \left(0.2346\right) \times 100 \approx 23.46\% \]

### Conclusion
The change in the kinetic energy of the electron is approximately **23.46%**.

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