Q.16 A ball of mass 330 g is moving with a constant speed, and its associated de Broglie wavelength is
1 × 10−33 m. The speed of the ball is ________ m s−1.
(h = 6.6 × 10−34 J s)
A ball of mass 330 g is moving with a constant speed, and its associated de Broglie wavelength is 1 × 10-33 m. The speed of the ball is ________ m s-1.
(h = 6.6 × 10-34 J s)
Final Answer: 2 m/s
Quick Solution
The de Broglie wavelength relates a particle’s wave-like properties to its momentum via the formula λ = h / p, where p = mv for non-relativistic speeds.
- Convert mass:
m = 330 g = 0.330 kg - Momentum:
p = h / λ = (6.6 × 10-34) / (1 × 10-33) = 0.66 kg m s-1 - Speed:
v = p / m = 0.66 / 0.330 = 2 m s-1
Step-by-Step Calculation
- De Broglie Equation:
λ = h / (mv)→ Rearrange:v = h / (mλ) - Substitute values:
v = (6.6 × 10-34) / (0.330 × 1 × 10-33) - Calculate:
v = (6.6 × 10-34) / (3.30 × 10-34) = 2 exactly - Result: Speed = 2 m/s
Introduction to De Broglie Wavelength Calculation
De Broglie wavelength calculation for a 330g ball moving at constant speed with λ=1×10-33m (h=6.6×10-34 Js) reveals wave-particle duality. This quantum physics concept, essential for JEE Main/Advanced, yields speed of 2 m/s via λ=h/(mv).
De Broglie Formula Derivation
Louis de Broglie proposed λ = h / p where p=mv (momentum). For macroscopic objects like 330g ball, wavelength is extremely small (~10-33m), making quantum effects negligible in daily observations but precisely calculable.
Why Such Small Wavelength for Macroscopic Ball?
- Huge momentum: 330g × 2 m/s = significant momentum
- Tiny wavelength: λ ∝ 1/(mass × speed)
- No observable interference: Unlike electrons in experiments
- Perfect for JEE/NEET: Illustrates quantum limits
Key Takeaways for JEE/NEET Students
- Always convert mass to kg for SI units
- De Broglie:
λ = h/(mv)→v = h/(mλ) - Macroscopic objects have negligible quantum effects
- Answer: 2 m/s (numerical type question)
