Question 19: Dual Nature of Matter and Radiation

Question

The mass of a particle $$\mathrm{A}$$ is double that of the particle $$\mathrm{B}$$ and the kinetic energy of $$\mathrm{B}$$ is $$\frac{1}{8}$$th that of A then the ratio of the de- Broglie wavelength of A to that of B is:

Accepted Answer

To solve this problem, let's start by understanding the relationship between the kinetic energy, mass, and de Broglie wavelength of the particles.
The de Broglie wavelength $$\lambda$$ of a particle is given by:
$$ \lambda = \frac{h}{p} $$
where $ h $ is Planck's constant and $ p $ is the momentum of the particle.
The momentum $ p $ can be expressed as:
$$ p = \sqrt{2mK} $$
where $ m $ is the mass and $ K $ is the kinetic energy of the particle.
Let's denote the masses and kinetic energies of particles $$\mathrm{A}$$ and $$\mathrm{B}$$ as follows:
Mass of particle $$\mathrm{A}$$: $$ m_A $$
Mass of particle $$\mathrm{B}$$: $$ m_B $$
Since the mass of particle $$\mathrm{A}$$ is double that of particle $$\mathrm{B}$$:
$$ m_A = 2m_B $$
Kinetic energy of particle $$\mathrm{A}$$: $$ K_A $$
Kinetic energy of particle $$\mathrm{B}$$: $$ K_B $$
Given that the kinetic energy of particle $$\mathrm{B}$$ is $$ \frac{1}{8} $$th that of $$\mathrm{A}$$:
$$ K_B = \frac{K_A}{8} $$
Now, let's find the ratio of the de Broglie wavelengths of particles $$\mathrm{A}$$ and $$\mathrm{B}$$. The de Broglie wavelength for particle $$\mathrm{A}$$ is:
$$ \lambda_A = \frac{h}{\sqrt{2m_A K_A}} $$
The de Broglie wavelength for particle $$\mathrm{B}$$ is:
$$ \lambda_B = \frac{h}{\sqrt{2m_B K_B}} $$
We need to find the ratio $$ \frac{\lambda_A}{\lambda_B} $$:
$$ \frac{\lambda_A}{\lambda_B} = \frac{\frac{h}{\sqrt{2m_A K_A}}}{\frac{h}{\sqrt{2m_B K_B}}} = \frac{\sqrt{2m_B K_B}}{\sqrt{2m_A K_A}} = \sqrt{\frac{m_B K_B}{m_A K_A}} $$
Substitute the given values:
$$ \frac{\lambda_A}{\lambda_B} = \sqrt{\frac{m_B \cdot \frac{K_A}{8}}{2m_B \cdot K_A}} = \sqrt{\frac{\frac{m_B K_A}{8}}{2m_B K_A}} = \sqrt{\frac{1}{16}} = \frac{1}{4} $$
Thus, the ratio of the de Broglie wavelength of $$ \mathrm{A} $$ to that of $$ \mathrm{B} $$ is:
Option C: 1 : 4

Answer

1 : 2

Answer

2 : 1

Answer

1 : 4

Answer

4 : 1

Dual Nature of Matter and Radiation

The mass of a particle $\mathrm{A} is double that of the particle \mathrm{B} and the kinetic energy of \mathrm{B} is \frac{1}{8}$th that of A then the ratio of the de- Broglie wavelength of A to that of B is:

Choose the correct answer: