Login/Sign Up
Home » 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.)
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.)
CBSE | Class 12 | NCERT | Nuclei | Physics
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.)

Answer –

The distance between the centres of two deuterons, d, is calculated as follows:

1st deuteron radius + 2nd deuteron radius

Radius of a deuteron nucleus = 2 f m = 2 × 10−15 m

d = 2 × 10−15 + 2 × 10−15 = 4 × 10−15 m

Charge on a deuteron nucleus is given as follows  –

Charge on an electron = e = 1.6 × 10−19 C

The two-deuteron system’s potential energy is:

v=\frac{{{e}^{2}}}{4\pi {{\in }_{0}}d}

Where,                                                                  

ϵ0​ is the Permittivity of free space, and

\frac{1}{4\pi {{\in }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}

Therefore,

V=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{4\times {{10}^{-15}}}J=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{4\times {{10}^{-15}}\times \left( 1.6\times {{10}^{-19}} \right)}eV

Therefore, V= 360 k eV

Hence, the height of the potential barrier of the two-deuteron system is 360 k eV.

i

More Material