Question 51: Atoms & Nuclei

Question

A particle at rest decays in to two particles of mass $$m_1$$ and $$m_2$$ and move with velocities $$v_1$$ and $$v_2$$. The ratio of their de Broglie wave length $$\frac{\lambda_1}{\lambda_2}$$ is:

Accepted Answer

The de Broglie wavelength of a particle is given by the formula:
$$\lambda = \frac{h}{p},$$
where $h$ is Planck's constant and $p$ is the momentum of the particle. The momentum $p$ can be expressed in terms of the mass $m$ and velocity $v$ of the particle as $p = m \cdot v$.
For the two particles in question, their de Broglie wavelengths are given by:
$$\lambda_1 = \frac{h}{p_1} = \frac{h}{m_1 v_1},$$
$$\lambda_2 = \frac{h}{p_2} = \frac{h}{m_2 v_2}.$$
To find the ratio $\frac{\lambda_1}{\lambda_2}$, we divide the equations:
$$\frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{m_1 v_1}}{\frac{h}{m_2 v_2}} = \frac{m_2 v_2}{m_1 v_1}.$$
Given that the original particle is at rest before decaying into two particles, by the conservation of momentum, the total momentum before decay must equal the total momentum after decay. Before the decay, the momentum is 0 (since the particle is at rest), and after the decay, it must still sum to 0. Thus, we have:
$$m_1v_1 = m_2v_2.$$
Substituting $m_1v_1 = m_2v_2$ into the equation for the ratio of the de Broglie wavelengths, we get:
$$\frac{\lambda_1}{\lambda_2} = \frac{m_2 v_2}{m_1 v_1} = 1.$$
Therefore, the ratio of their de Broglie wavelengths $\frac{\lambda_1}{\lambda_2}$ is 1 : 1. The correct option is:
Option B 1 : 1

Answer

1 : 4

Answer

1 : 1

Answer

1 : 2

Answer

2 : 1

Atoms & Nuclei

A particle at rest decays in to two particles of mass $m_1 and m_2 and move with velocities v_1 and v_2. The ratio of their de Broglie wave length \frac{\lambda_1}{\lambda_2}$ is:

Choose the correct answer: