Question

The de Broglie wavelength associated with a particle of momentum \(p\) is

• A. \(\lambda=\frac{h}{p}\)
• B. \(\lambda=hp\)
• C. \(\lambda=\frac{p}{h}\)
• D. \(\lambda=\frac{h}{mv^2}\)

Hint:

The de Broglie wavelength of a moving particle is inversely proportional to its momentum.

Answer 1

The Correct Option is A

Concept:
According to de Broglie hypothesis, every moving particle has a wave associated with it.

Step 1: The de Broglie wavelength is given by: \[ \lambda=\frac{h}{p} \]

Step 2: Here: \[ h=\text{Planck's constant} \] and \[ p=\text{momentum of particle} \]

Step 3: If the particle has mass \(m\) and velocity \(v\), then: \[ p=mv \]

Step 4: Therefore: \[ \lambda=\frac{h}{mv} \] \[ \boxed{\lambda=\frac{h}{p}} \]