Question

The de Broglie wavelength of the matter wave associated with an object dropped from a height \(x\), when it reaches the ground is proportional to

• A. \(x^3\)
• B. \(\dfrac{1}{\sqrt{x}}\)
• C. \(\sqrt{x}\)
• D. \(x^{3/2}\)
• E. \(x\)

Hint:

If kinetic energy depends on height, first find velocity from energy conservation, then use \[ \lambda=\frac{h}{mv} \]

Answer 1

The Correct Option is B

For a body dropped from height \(x\), \[ mgh=\frac12 mv^2 \] So, \[ v=\sqrt{2gx} \] Momentum: \[ p=mv \propto \sqrt{x} \] de Broglie wavelength: \[ \lambda=\frac{h}{p} \] Therefore, \[ \lambda \propto \frac{1}{\sqrt{x}} \] Hence, \[ \boxed{(B)\ \frac{1}{\sqrt{x}}} \]