The de Broglie wavelength and the kinetic energy of a particle are 2000 Å and 1 eV respectively. If its kinetic energy becomes 1 MeV, its de Broglie wavelength becomes
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 2\,\mathrm{\mathring{A}} \)
For a particle of mass \( m \) and kinetic energy \( K \), the momentum is \( p=\sqrt{2mK} \), so the de Broglie wavelength is \( \lambda=h/\sqrt{2mK} \). For the same particle, \( \lambda \) is inversely proportional to the square root of the kinetic energy.
\( \begin{aligned} \frac{\lambda_2}{\lambda_1} &= \sqrt{\frac{K_1}{K_2}} \\ &= \sqrt{\frac{1\,\mathrm{eV}}{10^{6}\,\mathrm{eV}}} \\ &= 10^{-3} \end{aligned} \)
The new wavelength is:
\( \begin{aligned} \lambda_2 &= 10^{-3}\times\lambda_1 \\ &= 10^{-3}\times2000\,\mathrm{\mathring{A}} \\ &= 2\,\mathrm{\mathring{A}} \end{aligned} \)