The de Broglie wavelength associated with a particle A moving with kinetic energy K, is λ. The de Broglie wavelength associated with another particle of mass half that of particle A and moving with kinetic energy 4 K will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.70 λ
The de Broglie wavelength is \( \lambda = h/p \), and a particle of mass m and kinetic energy K has momentum \( p = \sqrt{2mK} \).
\( \displaystyle \lambda = \frac{h}{\sqrt{2mK}} \)
For the second particle the mass is m/2 and the kinetic energy is 4K:
\( \displaystyle \lambda' = \frac{h}{\sqrt{2 \times \frac{m}{2} \times 4K}} = \frac{h}{\sqrt{4mK}} \)
\( \displaystyle \frac{\lambda'}{\lambda} = \sqrt{\frac{2mK}{4mK}} = \frac{1}{\sqrt{2}} \)
\( \displaystyle \lambda' = \frac{\lambda}{\sqrt{2}} \approx 0.707\,\lambda \approx 0.70\,\lambda \)