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The gravitational force between a H-atom and another particle of mass m will be given by Newton’s law: F=GM.mr2, where r is in km and ______. - Physics

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Question

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 ______.

Options

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

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

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

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

MCQ
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Solution

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

Explanation:

The gravitational force between an H-atom and another
particle of mass m will be given by Newton’s law, F = G M.m/r2

Here M is the effective mass of the Hydrogen atom.

Let us learn how to find the effective mass of a Hydrogen atom.
Suppose you start with a proton and an electron separated by a large distance. The mass of this system is just `m_("proton") + m_("electron")`.

Now let the proton and electron fall towards each other under their mutual electrostatic attraction. As they fall they will speed up, so by the time the proton and electron are about one hydrogen atom radius apart they are moving at a high speed. Note that we haven’t added or removed any energy, so the mass/energy of the system is still `m_("proton") + m_("electron")`.

The trouble is that this would not form a hydrogen atom because the proton and electron will just speed past each other and fly away again. To form a hydrogen atom we have to take the kinetic energy of the electron and proton out of the system so we can bring them to a stop. Let’s call the kinetic energy Ek. This energy has a mass given by Einstein’s famous equation E = mc2, so the dying mass of our atom is the mass we started with less than the energy we’ve taken out:

`M = m_("proton") + m_("electron") - E_k/c^2`

And Ek is just the binding energy. That's why we have ta subtract off the binding energy. Hence the effective mass of the atom can be written as `M = m_("proton") + m_("electron") - B/c^2`.

Or you could look at the problem the other way around. Start with a hydrogen atom of mass `m_("proton") + m_("electron")`. To split the electron and proton apart we have to add energy. In fact the amount of energy we have to add is just the binding energy, and this adds a mass of `B/c^2`.

Having split the atom we now just have a separate proton and electron, with a combined mass of `m_("proton") + m_("electron")`, so we have: `M + B/c^2 = m_("proton") + m_("electron")` and a quick rearrangement once again gives us: `M = m_("proton") + m_("electrons") - B/c^2`

From above discussion it is clear that Effective mass of hydrogen atom `M = m_("proton") + m_("electrons") - B/c^2`, where B is BE of hydrogen atom = 13.6 eV.

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Nuclear Force
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Chapter 13: Nuclei - Exercises [Page 81]

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NCERT Exemplar Physics [English] Class 12
Chapter 13 Nuclei
Exercises | Q 13.02 | Page 81
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