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Q. 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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For a head-on collision of two deuterons, the closest  distances between their centres will be d=2\timesr

d=2\times2.0

d=4.0 fm

d=4\times10-15 m

charge on each deuteron = charge of one proton=q =1.6\times10-19 C

The maximum electrostatic potential energy of the system during the head-on collision will be E

\\=\frac{q^{2}}{4\pi \epsilon _{0}d}\\ =\frac{9\times 10^{9}\times (1.6\times 10^{-19})^{2}}{4\times 10^{-15}}\ J\\ =\frac{9\times 10^{9}\times (1.6\times 10^{-19})^{2}}{4\times 10^{-15}\times 1.6\times 10^{-19}}\ eV\\=360\ keV

The above basically means to bring two deuterons from infinity to each other would require 360 keV of work to be done or would require 360 keV of energy to be spent.

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Sayak

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