Asked by lindsay
to start a nuclear fusion reaction, two hydrogen atoms of charge 1.602x10^-19 and mass 1.67x10^-27kg must be fired at each other. If each particle has an initial velocity of 2.7x10^6m/s when released, what is their minimum seperation?
Answers
Answered by
Elena
KE =PE
mv²/2= ke²/r ,
r= 2•k•e²/m•v²,
where
k =9•10^9 N•m²/C²,
e =1.6•10^-19 C,
m =1.66•10^-27 kg.
mv²/2= ke²/r ,
r= 2•k•e²/m•v²,
where
k =9•10^9 N•m²/C²,
e =1.6•10^-19 C,
m =1.66•10^-27 kg.
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