Question

A deuteron 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

Using conservation of energy, change in electrostatic energy equals the kinetic energy gained.

$E=\frac{1}{2}m{v}^2=qV$

$p=mv=\sqrt{2mqV}$

1)

De-broglie wavelength is given by: $\lambda =\frac{h}{p}$

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

$\frac{{\lambda }_d}{{\lambda }_{\alpha }}=\frac{\sqrt{m_{\alpha }{q}_{\alpha }}}{\sqrt{m_d{q}_d}}$

$\frac{{\lambda }_d}{{\lambda }_{alpha}}=\frac{\sqrt{4\times 2}}{\sqrt{2\times 1}}=2$

Hence, wavelength of deuteron is more than the wavelength of the alpha particle.

2)

Kinetic energy, $E=qV$ as discussed above.

$\frac{E_d}{E_{\alpha }}=\frac{q_d}{q_{\alpha }}=\frac{1}{2}$

Hence, kinetic energy of alpha particle is less than the kinetic energy of deuteron.