If numerical value of mass and velocity are equal then de Broglie wavelength in terms of K.E. is

Options

A. `(mh)/(2K.E.)`
B. `(vh)/(2K.E)`
C. Both are correct
D. None is correct

Explanation

To solve the problem, we need to derive the de Broglie wavelength in terms of kinetic energy given that the numerical values of mass (m) and velocity (v) are equal.

### Step-by-step Solution:

1. **Understanding de Broglie Wavelength**:
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is the Planck constant and \( p \) is the momentum.

2. **Expressing Momentum**:
Momentum (p) is defined as:
\[ p = mv \]
Therefore, we can substitute this into the de Broglie wavelength equation:
\[ \lambda = \frac{h}{mv} \]

3. **Relating Kinetic Energy**:
The kinetic energy (K.E) of an object is given by:
\[ K.E = \frac{1}{2} mv^2 \]
From this equation, we can express \( mv^2 \) in terms of kinetic energy:
\[ mv^2 = 2 \cdot K.E \]

4. **Substituting into Wavelength Formula**:
Now, we can substitute \( mv^2 \) into the de Broglie wavelength equation. To do this, we multiply and divide \( \lambda \) by \( v \):
\[ \lambda = \frac{h}{mv} \cdot \frac{v}{v} = \frac{hv}{mv^2} \]
Now, substituting \( mv^2 \) with \( 2 \cdot K.E \):
\[ \lambda = \frac{hv}{2 \cdot K.E} \]

5. **Using the Condition \( m = v \)**:
Since the problem states that the numerical values of mass and velocity are equal, we can replace \( v \) with \( m \):
\[ \lambda = \frac{hm}{2 \cdot K.E} \]

6. **Final Expression**:
Thus, we have derived the de Broglie wavelength in terms of kinetic energy:
\[ \lambda = \frac{hm}{2 \cdot K.E} \]

### Conclusion:
The de Broglie wavelength in terms of kinetic energy, given that mass and velocity are equal, is:
\[ \lambda = \frac{hm}{2 \cdot K.E} \]