Question

A particle of mass \(M\) at rest splits up into two particles of masses \(m_1\) and \(m_2\) having non-zero velocities. Calculate the ratio of the de Broglie wavelengths associated with the two particles.

Hint:

Whenever a stationary object splits purely due to internal forces into two fragments, they will always fly away with equal and opposite momentum magnitudes. Since de Broglie wavelength depends solely on momentum (\(\lambda = h/p\)), their wavelengths must be exactly equal, regardless of their different masses!

Answer 1

Concept: This problem involves two distinct physical principles:
• The Law of Conservation of Linear Momentum: In the total absence of any external net unbalanced force acting upon a physical system, the total initial linear momentum (\(\vec{P}_{\text{initial}}\)) of the system must equal its total final linear momentum (\(\vec{P}_{\text{final}}\)).
• The de Broglie Wavelength Hypothesis: Every moving material particle exhibits dual wave-particle characteristics. The de Broglie wavelength (\(\lambda\)) associated with a moving matter particle is inversely proportional to its linear momentum magnitude (\(p\)), expressed by: \[ \lambda = \frac{h}{p} \] where \(h\) is Planck's constant (\(h \approx 6.626 \times 10^{-34} \text{ J}\cdot\text{s}\)).

Step 1: Applying Conservation of Linear Momentum
Let us establish the system parameters. Initially, we have a single isolated stationary particle of mass \(M\). Because it is explicitly stated to be at rest, its initial velocity vector is zero (\(\vec{v}_0 = \vec{0}\)). Therefore, the total initial linear momentum of the system is: \[ \vec{P}_{\text{initial}} = M \cdot \vec{v}_0 = M \cdot \vec{0} = \vec{0} \] The particle spontaneously splits or undergoes an internal explosion, dividing into two distinct fragments of mass \(m_1\) and \(m_2\). Let their resulting final velocity vectors immediately following the split be denoted as \(\vec{v}_1\) and \(\vec{v}_2\) respectively. The total final linear momentum of the two-particle system is the vector sum of their individual momenta: \[ \vec{P}_{\text{final}} = \vec{p}_1 + \vec{p}_2 = m_1\vec{v}_1 + m_2\vec{v}_2 \] Since no external forces acted on the system during the internal splitting process, linear momentum is conserved perfectly: \[ \vec{P}_{\text{initial}} = \vec{P}_{\text{final}} \] \[ \vec{0} = \vec{p}_1 + \vec{p}_2 \] Rearranging this vector equation yields: \[ \vec{p}_1 = -\vec{p}_2 \] This vector relationship physically means that the linear momentum vector of the first particle is exactly equal in magnitude and directly opposite in spatial direction to the linear momentum vector of the second particle. Taking the absolute mathematical magnitude (norm) of both sides of this equation: \[ |\vec{p}_1| = |-\vec{p}_2| \quad \Rightarrow \quad p_1 = p_2 \] Thus, the scalar magnitudes of the linear momenta of both fragments are perfectly identical.

Step 2: Determining the ratio of the de Broglie wavelengths
According to the de Broglie relation, the wavelength associated with the first particle of mass \(m_1\) moving with momentum magnitude \(p_1\) is: \[ \lambda_1 = \frac{h}{p_1} \] Similarly, the wavelength associated with the second particle of mass \(m_2\) moving with momentum magnitude \(p_2\) is: \[ \lambda_2 = \frac{h}{p_2} \] Now, let us form the mathematical ratio of the first wavelength to the second wavelength (\(\frac{\lambda_1}{\lambda_2}\)): \[ \frac{\lambda_1}{\lambda_2} = \frac{\left(\frac{h}{p_1}\right)}{\left(\frac{h}{p_2}\right)} \] Since Planck's constant \(h\) is an identical universal scalar factor in both the numerator and the denominator, it cancels out completely: \[ \frac{\lambda_1}{\lambda_2} = \frac{p_2}{p_1} \] From Step 1, the conservation of linear momentum proved that the momentum magnitudes are identical (\(p_1 = p_2\)). Substituting this equality into our ratio gives: \[ \frac{\lambda_1}{\lambda_2} = \frac{p_1}{p_1} = 1 \] Therefore, the ratio of their de Broglie wavelengths is: \[ \lambda_1 : \lambda_2 = 1 : 1 \] The wavelengths are equal, which corresponds exactly to option (C).