A dbuteron and an alpha particle are accelerated with the same ac… - Vidyadip

Question

A dbuteron and an alpha particle are accelerated with the same accelerating potential. Which one of the two has:

Greater value of de-Broglie wavelength, associated with it and
Less kinetic energy? Explain.

✓

Answer

deutron

$\lambda = \frac{\text{h}}{\sqrt{2\text{mqV}}}$

$\frac{\lambda_d}{\lambda_{\alpha}}=\sqrt{\frac{m_\alpha{q_\alpha}}{m_d{q_d}}}=\sqrt{\frac{{2m_d}\times{2q_d}}{m_d{q_d}}}$

$=\frac{2}{1}$

$\Rightarrow\lambda_{d} > \lambda_{\alpha}$

deuteron

KE = qV

$\therefore$ $q_{\alpha}>q_{d}$

for the same accelerating potential, we have

$(KE)_d>(KE){\alpha}$

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