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The beryllium isotope 4Be is unstable and decays into two α particles (helium nuclei of mass 6.68 × 10^−27 kg) with the release of 1.5 × 10^−14 J of energy. Determine the velocities of the two α particles that arise from the decay of a 4Be nucleus at rest, assuming that all the energy appears as kinetic energy of the particles. Answer in units of m/s.

Answer :

Answer:

v=1.56 x 10⁶ m/s

Explanation:

Given that

Mass alpha ,m=6.68 x 10⁻²⁷ kg

4Be is unstable

4Be  → 2α   + 1.5 × 10⁻¹⁴ J

Lets take speed of α particle is v m/s

So from energy conservation

[tex]2.k_\alpha =2.\dfrac{1}{2}m_\alpha v^2=m_\alpha v^2[/tex]

[tex]E=m_\alpha v^2[/tex]

Here

E= 1.5 × 10⁻¹⁴ J

m=6.68 x 10⁻²⁷ kg

1.5 × 10⁻¹⁴  = 6.68 x 10⁻²⁷ x v²

v²= 0.224 x 10¹³

v=1.56 x 10⁶ m/s

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