If kinetic energy of an electron is increases by 69% then what is the percentage in its de-broglie wavelength :-

Options

A. Decreases by 23.07%
B. Decreases by 15%
C. Increases by 40%
D. Decreases by 70%

Answer

To solve the problem, we need to analyze the relationship between the kinetic energy of an electron and its de Broglie wavelength. Here’s a step-by-step solution:

### Step 1: Understand the relationship between kinetic energy and velocity
The kinetic energy (KE) of an electron is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron and \( v \) is its velocity.

### Step 2: Calculate the new kinetic energy after a 69% increase
If the kinetic energy increases by 69%, the new kinetic energy \( KE' \) can be expressed as:
\[ KE' = KE + 0.69 \times KE = 1.69 \times KE \]

### Step 3: Relate the new kinetic energy to the new velocity
Substituting the expression for kinetic energy into the equation, we have:
\[ KE' = \frac{1}{2} mv'^2 = 1.69 \times \frac{1}{2} mv^2 \]
This simplifies to:
\[ v'^2 = 1.69 v^2 \]

### Step 4: Solve for the new velocity
Taking the square root of both sides gives us:
\[ v' = \sqrt{1.69} \times v = 1.3 v \]
This indicates that the new velocity \( v' \) is 1.3 times the initial velocity \( v \).

### Step 5: Write the de Broglie wavelength formula
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant.

### Step 6: Calculate the new de Broglie wavelength
The new de Broglie wavelength \( \lambda' \) with the new velocity \( v' \) is:
\[ \lambda' = \frac{h}{mv'} = \frac{h}{m(1.3v)} = \frac{1}{1.3} \cdot \frac{h}{mv} = \frac{\lambda}{1.3} \]

### Step 7: Determine the percentage change in wavelength
The ratio of the new wavelength to the old wavelength is:
\[ \frac{\lambda'}{\lambda} = \frac{1}{1.3} \approx 0.769 \]
This means the new wavelength is approximately 76.9% of the original wavelength.

### Step 8: Calculate the decrease in wavelength
The decrease in wavelength can be calculated as:
\[ \text{Decrease} = 1 - 0.769 = 0.231 \]
To find the percentage decrease:
\[ \text{Percentage decrease} = 0.231 \times 100 \approx 23.1\% \]

### Final Answer
Thus, the percentage decrease in the de Broglie wavelength is approximately **23%**.

---