Question

A proton is accelerated from rest through a potential difference of 500 volts. Find its final momentum. [m_p = 1.67 × 10^-27 kg, e = 1.6 × 10^-19 C]

Answer 1

Data : m_p = 1.67 × 10^-27 kg, 

e = 1.6 × 10^-19 C, 

u = 0, V = 500 V

Initial KE, KE; = \(\cfrac12\) m_pu² = 0

∴ ∆KE = KE_f – KE_i = KE_f

∆KE = qV

∴ KE_f = qV KE_f = \(\cfrac12\) m_pv² = \(\cfrac{p^2_1}{2m_p}\)

where p_f = m_pv ≡ the magnitude of the final momentum of the proton.

∴ P_f² = 2 m_pqV

∴ P_f = \(\sqrt{2m_pqV}\)

= \(\sqrt{2(1.67\times10^{-27}\,kg)(1.6\times10^{-19}\,C)(500\,V)}\)

= 5.169 × 10^-22 kg∙m/s 

The momentum is directed along the applied electric field.