Question 12 of 13advanced🔧 ApplyNumerical3 marks

Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)

Correct Answer

The height of the potential barrier for a head-on collision of two deuterons is 360 keV.

Exercise: EXERCISES | Q: 13.9 | (Page 17)
For More Understanding

Explanation

The question asks to calculate the potential barrier height which is the Coulomb repulsion energy when two deuterons just touch each other. The context provides the hint that deuterons can be taken as hard spheres of radius 2.0 fm. When two deuterons just touch, the distance between their centers equals twice the radius. The Coulomb potential energy at this distance gives the barrier height. The context also mentions that barrier height for two protons is ~400 keV, which serves as a reference point - our calculated value of 360 keV for deuterons is reasonable since deuterons have the same charge (+e) as protons but larger radius.

Solution Steps

  1. Step 1: Given: Radius of deuteron, r = 2.0 fm = 2.0×102.0 \times 10⁻¹⁵ m. Charge on each deuteron = +e = 1.6×101.6 \times 10⁻¹⁹ C.

  2. Step 2: When two deuterons just touch each other, distance between their centres: d = 2r = 2×2.02 \times 2.0 fm = 4.0 fm = 4.0×104.0 \times 10⁻¹⁵ m.

  3. Step 3: The height of potential barrier is given by Coulomb potential energy: U = (1/4πε₀) × e²/d

  4. Step 4: Substituting values: U = (9×109 \times 10⁹) × (1.6×101.6 \times 10⁻¹⁹)² / (4.0×104.0 \times 10⁻¹⁵) = 5.76×105.76 \times 10⁻¹⁴ J

  5. Step 5: Converting to keV: U = 5.76×105.76 \times 10⁻¹⁴ / (1.6×101.6 \times 10⁻¹⁶) keV = 360 keV

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