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JEE MainPhysicsNuclear Physics

Let E_b be the nuclear binding energy of an alpha particle ( ₂^4 He ). If m_p and m_n represent the masses of a proton and a neutron respectively, and c is the speed of light in vacuum, then the mass of the alpha particle nucleus is given by:

Options

  1. A2m_p + 2m_n - E_b c^2
  2. B2m_p + 2m_n + E_b c^2
  3. C4m_p + 4m_n - E_b c^2
  4. D2m_p + 4m_n - E_b c^2

Correct answer

A. 2m_p + 2m_n - E_b c^2

Step-by-step solution

The binding energy E_b is related to the mass defect m by the equation: E_b = m c^2 The mass defect for an alpha particle ( ₂^4 He ), which consists of 2 protons and 2 neutrons, is: m = 2m_p + 2m_n - M_ He Substituting this into the binding energy equation gives: E_b = (2m_p + 2m_n - M_ He ) c^2 Rearranging to solve for the mass of the nucleus M_ He : M_ He = 2m_p + 2m_n - E_b c^2 Answer: 2m_p + 2m_n - E_b c^2

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