Question

If the kinetic energy of a particle having wavelength \(x \, {\AA}\) is increased to three times, its de Broglie wavelength (in {\AA}) is:

• A. \(3x\)
• B. \(\sqrt{3} x\)
• C. \(\frac{x}{\sqrt{3}}\)
• D. \(\frac{x}{3}\)

Hint:

Kinetic energy and momentum are related such that increasing kinetic energy decreases the de Broglie wavelength by the square root of the increase factor.

Answer 1

The Correct Option is C

The de Broglie wavelength \(\lambda\) is related to momentum \(p\) by: \[ \lambda = \frac{h}{p} \] The kinetic energy \(K.E.\) is: \[ K.E. = \frac{p^2}{2m} \implies p = \sqrt{2mK.E.} \] If kinetic energy is increased to three times, then: \[ K.E._{new} = 3 \times K.E. \] Corresponding momentum: \[ p_{new} = \sqrt{2m \times 3 K.E.} = \sqrt{3} \times p \] Therefore, new wavelength \(\lambda_{new}\): \[ \lambda_{new} = \frac{h}{p_{new}} = \frac{h}{\sqrt{3} p} = \frac{\lambda}{\sqrt{3}} \] Given the original wavelength is \(x\), so: \[ \lambda_{new} = \frac{x}{\sqrt{3}} \]