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Physics Ncert Solutions Class 12th Physics Class 12th Nuclei

13.20 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.)

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Payal Gupta | Contributor-Level 10

3 months ago

13.20 When two deuterons collide head-on, the distance between their centers, d is given as

d = Radius of 1st deuteron + Radius of 2nd deuteron

Radius of the deuteron nucleus = 2 fm = 2 $\times 10^{- 15}$ m

Hence d = 2 $\times 10^{- 15} + 2 \times 10^{- 15}$ = 4 $\times 10^{- 15}$ m

Charge on a deuteron nucleus = Charge on an electron = 1.6 $\times 10^{- 19}$ C

Potential energy of two-deuteron system: $V$ = $\frac{e^{2}}{4 π ?_{0} d}$

Where $?_{0}$ = permittivity of free space

$\frac{1}{4 π ?_{0}}$ = 9 $\times 10^{9}$ N $m^{2} C^{- 1}$

V = $\frac{{(1.6 \times 10^{- 19})}^{2} \times 9 \times 10^{9}}{4 \times 10^{- 15}}$ J = $\frac{{(1.6 \times 10^{- 19})}^{2} \times 9 \times 10^{9}}{4 \times 10^{- 15} \times 1.6 \times 10^{- 19}}$ eV = 360 $\times 10^{3}$
eV = 360 keV

Hence the height of the potential barrier of the two-deuteron system is 360 keV.

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13.20 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.)

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