Question

A particle of mass $M$ at rest decays into two masses $$m_1$$and $$m_2$$with non-zero velocities. What is the ratio of $\frac{{\lambda }_1}{{\lambda }_2}$of de Broglie wavelengths of particles?

Answer 1

Step 1. Formula used:
$$p={\displaystyle \frac{hc}{\lambda }}$$
Here, $h$ is Planck’s constant.

$c$ is the speed of light.

$$\lambda$$ is the de Broglie wavelength.

Step 2: Formula Used

Linear momentum of the particle mass $m$ moving with velocity $v$,
$$p=mv$$
$p$ is the linear momentum of the particle.
Step 3. Calculating the ratio of de Broglie wavelength:
We know that, according to de Broglie's hypothesis, the momentum of the particle is,
$$p={\displaystyle \frac{hc}{\lambda }}$$
Here, $h$ is Planck’s constant, $c$ is the speed of light,

$$\lambda$$ is the de Broglie wavelength.
According to the law of conservation of momentum,
We can write,
$$Mv=m_1{v}_1+m_2{v}_2$$ $v_2$
$v$ is the velocity of the parent particle,

$$v_1$$ is the velocity $$m_1$$

$v_2$ is the velocity of $$m_2$$.

Since the parent particle with mass $M$ remains at rest, the initial velocity $v$ is zero.

Therefore, the above equation becomes,
$$0=m_1{v}_1+m_2{v}_2$$
$$\Rightarrow m_1{v}_1=-m_2{v}_2$$
The particle of mass $$m_1$$and particle of mass $$m_2$$has equal momentum.
So, we can write,
Since, $|p_1|=\left|p_2\right|$
$$\Rightarrow {\displaystyle \frac{hc}{{\lambda }_1}}={\displaystyle \frac{hc}{{\lambda }_2}}$$
The wavelength of these particles is equal,
$$\therefore {\displaystyle \frac{{\lambda }_1}{{\lambda }_2}}=1$$

Therefore, the ratio of de Broglie wavelength of particles is $1$.