Question
A dbuteron and an alpha particle are accelerated with the same accelerating potential. Which one of the two has:
  1. Greater value of de-Broglie wavelength, associated with it and
  2. Less kinetic energy? Explain.

Answer

  1. deutron
$\lambda = \frac{\text{h}}{\sqrt{2\text{mqV}}}$
$\frac{\lambda_d}{\lambda_{\alpha}}=\sqrt{\frac{m_\alpha{q_\alpha}}{m_d{q_d}}}=\sqrt{\frac{{2m_d}\times{2q_d}}{m_d{q_d}}}$
$=\frac{2}{1}$
$\Rightarrow\lambda_{d} > \lambda_{\alpha}$ 
  1. deuteron
KE = qV
$\therefore$ $q_{\alpha}>q_{d}$
for the same accelerating potential, we have
$(KE)_d>(KE){\alpha}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free