Practice portal › Atoms › Hydrogen Spectrum and Transitions
Asked in CBSE AIPMT 1994 · Links to recoil, photoelectric effect and X-rays
Given: both ions start from rest and fall through the same potential difference V.
Idea: the work done on the charge becomes kinetic energy, 1/2mv²=qV, so v=√(2qV)/m.
A doubly ionised helium atom has q=2e and m=4m_H; a hydrogen ion has q=e and m=m_H.
(v_He)/(v_H)=√(2e/4m_H)/(e/m_H)=√1/2.
So the ratio is 1/(√2): the heavier helium ion ends up the slower of the two.
Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.
Practise this with a timer