Home
Class 12
PHYSICS
The gravitational force between a H-atom...

The gravitational force between a H-atom and another particle of mass m will be given by Newton's law: `F=G (M.m)/(r^(2)`, where r is in km and

A

`M=m_("proton")+m_("electron ")`

B

`M=m_("proton")+m_("electron")-(B)/(c^(2))" "(B=13.6" eV")`.

C

M is not related to the mass of the hydrogen atom.

D

`M=m_("proton")+m_("electron")-(|V|)/(c^(2)) (|V|="magnitude of the potential energy of electron in the H-atom")`.

Text Solution

Verified by Experts

The correct Answer is:
B
During formation of H-atom, some mass of nucleons convert into energy by the equation
`E=mc^(2)`
This energy is used to bind the nucleons along with nucleus. So mass of atom becomes slightly less than sum of actual masses of nucleons and electrons.
Actual mass of H-atom `=M_(0)+M_(e)-("B.E.")/(c^(2))" "((B)/(c^(2))=" Binding energy")`
So, the binding energy of H atoms is 13.6 eV per atom.
Promotional Banner

Similar Questions

Explore conceptually related problems

The gravitational force between two point masses m_(1) and m_(2) at separation r is given by F = k(m_(1) m_(2))/(r^(2)) The constant k doesn't

The gravitional force between two point masses m_(1) and m_(2) at separation r is given by F=k (m_(1)m_(2))/r^(2) The constant k

Assertion : Mass of the rod AB is m_(1) and of particle P is m_(2) . Distance between centre of rod and particle is r. Then the gravitational force between the rod and the particle is F=(Gm_(1)m_(2))/(r^(2)) Reason : The relation F=(Gm_(1)m_(2))/(r^(2)) can be applied directly only to find force between two particles.

Gravitational force between two objects with mass m_(1)" and "m_(2) kept r distance apart is given as F_(G)= (Gm_(1)m_(2))/(r^2) , where G is gravitational constant. Express the relative error in F_(G) .

What is the dimensional formula of (a) volume (b) density and (c ) universal gravitational constant 'G' . The gravitational force of attraction between two objects of masses m and m separated by a distance d is given by F = (G m_(1) m_(2))/(d^(2)) Where G is the universal gravitational constant.

The gravitational force between two particles of masses m_(1) and m_(2) separeted by the same distance, then the gravitational force between them will be

Suppose that the gravitational force between two bodies separated by a distance R becomes inversely proportional to R (and not as (1)/(R^(2)) as given by Newton's law of gravitation). Then the speed of a particle moving in a circular orbit of radius R under such a force would be proportional to

Assertion : The centres of two cubes of masses m_(1) and m_(2) are separated by a distance r. The gravitational force between these two cubes will be (Gm_(1)m_(2))/(r^(2)) Reason : According to Newton's law of gravitation, gravitational force between two point masses m_(1) and m_(2) separated by a distance r is (Gm_(1)m_(2))/(r^(2)) .

If gravitational attraction between two points masses be given by F=G(m_(1)m_(2))/(r^(n)) . Then the period of a satellite in a circular orbit will be proportional to

Gravitational force, F=Gm_(1)m_(2)//r.