Two fast moving particles X and Y are associated with de Broglie wavelengths 1 nm and 4 nm respectively. If mass of X is nine times the mass of Y, then calculate ratio of kinetic energies of X and Y.

Explanation

To solve the problem, we need to calculate the ratio of the kinetic energies of two particles X and Y, given their de Broglie wavelengths and the relationship between their masses.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Wavelength of particle X, \( \lambda_x = 1 \, \text{nm} \)
- Wavelength of particle Y, \( \lambda_y = 4 \, \text{nm} \)
- Mass of particle X, \( m_x = 9 m_y \) (mass of X is nine times the mass of Y)

2. **Use the de Broglie Wavelength Formula:**
The de Broglie wavelength is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.

3. **Set Up the Equations for Both Particles:**
For particle X:
\[ \lambda_x = \frac{h}{m_x v_x} \implies 1 = \frac{h}{m_x v_x} \]
For particle Y:
\[ \lambda_y = \frac{h}{m_y v_y} \implies 4 = \frac{h}{m_y v_y} \]

4. **Take the Ratio of the Two Equations:**
\[ \frac{\lambda_x}{\lambda_y} = \frac{h/(m_x v_x)}{h/(m_y v_y)} \implies \frac{1}{4} = \frac{m_y v_y}{m_x v_x} \]

5. **Substituting the Mass Relationship:**
Since \( m_x = 9 m_y \):
\[ \frac{1}{4} = \frac{m_y v_y}{9 m_y v_x} \implies \frac{1}{4} = \frac{v_y}{9 v_x} \]

6. **Rearranging to Find the Velocity Ratio:**
\[ v_y = \frac{9}{4} v_x \]

7. **Kinetic Energy Formula:**
The kinetic energy (KE) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
Therefore, for particles X and Y:
\[ KE_x = \frac{1}{2} m_x v_x^2 \]
\[ KE_y = \frac{1}{2} m_y v_y^2 \]

8. **Finding the Ratio of Kinetic Energies:**
The ratio of the kinetic energies is:
\[ \frac{KE_x}{KE_y} = \frac{m_x v_x^2}{m_y v_y^2} \]
Substituting \( m_x = 9 m_y \) and \( v_y = \frac{9}{4} v_x \):
\[ \frac{KE_x}{KE_y} = \frac{9 m_y v_x^2}{m_y \left(\frac{9}{4} v_x\right)^2} \]
Simplifying:
\[ = \frac{9 v_x^2}{m_y \cdot \frac{81}{16} v_x^2} = \frac{9}{\frac{81}{16}} = \frac{9 \cdot 16}{81} = \frac{144}{81} = \frac{16}{9} \]

### Final Answer:
The ratio of the kinetic energies of particles X and Y is:
\[ \frac{KE_x}{KE_y} = \frac{16}{9} \]