Q.A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de Broglie wavelength of the particle varies cyclically between two values $\lambda_1, \lambda_2$ with $\lambda_1 > \lambda_2$. Which of the following statements are true?

Concept: De Broglie Wavelength

The de Broglie wavelength $\lambda = h/p$ is inversely proportional to the particle’s momentum $p$. Since the force is central (directed toward the origin), angular momentum is conserved.

Reasoning:

1. For a central force, the speed $v$ (and hence $p = mv$) is larger when the particle is closer to the origin (by conservation of energy: potential energy decreases, kinetic energy increases).
2. A smaller $\lambda$ means larger $p$, so the particle is nearer the origin when $\lambda = \lambda_2$ (the smaller wavelength). …

The key idea here is the de Broglie relation: $\lambda = h / p$, where $p$ is the magnitude of the particle's momentum. For a particle moving under a central force (always directed toward the origin), the orbit lies in a plane and obeys conservation of angular momentum. The speed — and therefore the momentum — changes as the distance from the origin changes. Let’s see why.

1. De Broglie wavelength and momentum

   Since $\lambda = h/p$, a larger $\lambda$ means smaller $p$, and a smaller $\lambda$ means larger $p$. So $\lambda_1 > \lambda_2$ tells us that when the wavelength is $\lambda_1$, the particle has less momentum than when it is $\lambda_2$.

2. Central force and orbital motion

   In a central force, angular momentum $L = m r v_\perp$ is conserved. Here $v_\perp$ is the component of velocity perpendicular to the radius vector. As the particle moves closer to the origin ($r$ decreases), $v_\perp$ must increase to keep $L$ constant. The total speed $v$ (and hence momentum $p = mv$) generally increases when $r$ decreases, because the perpendicular component dominates and the radial component is bounded.

   Tip

   For a bound orbit under an inverse-square law (like gravity), the speed is largest at the point closest to the focus (perihelion) and smallest at the farthest point (aphelion). This is a direct consequence of energy conservation: $E = \frac12 m v^2 - GMm/r$ is constant, so smaller $r$ gives larger $v$.

3. What kind of orbit gives cyclic variation?

   • Circular orbit: distance $r$ is constant, so speed is constant, momentum is constant, and de Broglie wavelength is constant. It cannot vary cyclically between two distinct values. So statement (A) is false. …

Method: Connecting Orbital Dynamics to a Physically Varying Quantity

When a quantity like de Broglie wavelength is said to vary cyclically as a particle orbits under a central force, the technique is to first translate the wavelength's variation into a variation of speed/momentum, then use standard orbital-mechanics facts about where speed is largest and smallest.

Steps

Step 1: Translate "wavelength varies" into "momentum (and hence speed) varies"

Since $\lambda_1 > \lambda_2$, the particle moves slower whenever its wavelength is $\lambda_1$ and faster whenever it is $\lambda_2$.

Step 2: Recall which orbit shapes actually allow speed to vary

For a central force, angular momentum about the origin is conserved, and (for a bound attractive orbit) total mechanical energy is conserved too. A circular orbit has constant radius, so speed is constant throughout — it cannot produce a wavelength that cycles between two different values. An elliptical orbit (with the force centre at a focus) has radius varying between a minimum and a maximum, so speed genuinely varies periodically.

Step 3: Use energy conservation to identify where speed is largest

$E = \frac12 mv^2 + U(r) = \text{constant}$ …