Calculate the height of the potential barrier | vedclass

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(D) The potential energy of the barrier at the point of contact is given by $U = k \frac{q_1 q_2}{r}$.Here, the charge of each deuteron is $q_1 = q_2

Vedclass · Accepted Answer

$5.76 	imes 10^{-14}$ $J$

Vedclass · Answer

(D) The potential energy of the barrier at the point of contact is given by $U = k \frac{q_1 q_2}{r}$.Here, the charge of each deuteron is $q_1 = q_2 = e = 1.6 	imes 10^{-19}$ $C$.The distance between the centers of the two deuterons at the point of contact is $r = R_1 + R_2 = 2 	ext{ fm} + 2 	ext{ fm} = 4 	ext{ fm} = 4 	imes 10^{-15}$ m.Substituting the values into the formula:$U = \frac{(9 	imes 10^9 	ext{ N m}^2/	ext{C}^2) 	imes (1.6 	imes 10^{-19} 	ext{ C})^2}{4 	imes 10^{-15} 	ext{ m}}$$U = \frac{9 	imes 10^9 	imes 2.56 	imes 10^{-38}}{4 	imes 10^{-15}}$$U = \frac{23.04 	imes 10^{-29}}{4 	imes 10^{-15}}$$U = 5.76 	imes 10^{-14}$ $J$.

Column-$I$	Column-$II$
$A$. Isotopes	$i$. $Li^7, Be^7$
$B$. Isobars	$ii$. $_8O^{18}, _9F^{19}$
$C$. Isotones	$iii$. $_1H^1, _1H^2$

Calculate the height of the potential barrier for a head-on collision of two deuterons. (Radius of deuteron is $2$ fm).

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