Question

The de Broglie wavelength is:

• A. \( \lambda = h m v \)
• B. \( \lambda = \frac{h}{m v} \)
• C. \( \lambda = h v \)
• D. \( \lambda = \frac{h v}{m} \)

Hint:

de Broglie wavelength relates particle momentum and wavelength via Planck's constant.

Answer 1

The Correct Option is B

According to de Broglie's hypothesis, every moving particle exhibits wave-like properties, with an associated wavelength known as the de Broglie wavelength. For a particle having momentum \( p = m v \), where \( m \) is its mass and \( v \) its velocity, the de Broglie wavelength \( \lambda \) is given by: \[ \lambda = \frac{h}{p} = \frac{h}{m v}, \] where \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J·s}\)). This relationship shows that the wavelength is inversely proportional to the momentum of the particle, linking particle and wave nature.