Final answer:
The ratio of the total acceleration of an electron in a singly ionized helium atom to that in a hydrogen atom, both in the ground state, is 16:1 according to Bohr's model, due to the helium ion having a doubled nuclear charge and a radius half of the Bohr radius.
Step-by-step explanation:
The question asks for the ratio of the total acceleration of an electron in a singly ionized helium atom (He+) compared to that in a hydrogen atom (H), both in the ground state. By applying Bohr's model we can analyze both systems.
In Bohr's model, the force acting on the electron is the centripetal force required to keep the electron in its circular orbit. This force is provided by the electrostatic force between the positively charged nucleus and the negatively charged electron. The electrostatic force is proportional to the product of the involved charges and inversely proportional to the square of the radius of the orbit. The acceleration 'a' of the electron is then governed by the relation F = ma, where 'm' is the mass of the electron and 'a' is its acceleration.
For a hydrogen atom, this force is Coulomb's law F = k(e^2/r2), where e is the charge of an electron, r is the radius of the orbit (which is the Bohr radius for the ground state of hydrogen), and k is Coulomb's constant. For He+, the nucleus has a charge of +2e, and so the electrostatic force is F = k(2e^2/r'2), where r' is the radius of the orbit for He+ in the ground state, which is half the Bohr radius (r' = r/2). The total acceleration 'a' for an electron in hydrogen is k(e^2/mr2), and for He+ it is k(4e^2/mr'2) which is k(16e^2/mr2) when expressed in terms of the Bohr radius 'r' for hydrogen.
Thus, the ratio of the accelerations is (16e^2/mr2) / (e^2/mr2) which simplifies to 16:1. However, with a typographical error in the question which mentioned the ratio as 4:1, the correct ratio of accelerations is instead 16:1.