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wlad13 [49]
3 years ago
13

"what potential difference is needed to give a helium nucleus (q=3.2×10^−19c) 140 kev of kinetic energy?"

Physics
1 answer:
omeli [17]3 years ago
3 0

whenever a charge particle is moving under the potential difference at that time the kinetic energy gained by the charged particle is given as

            K.E=\frac{1}{2} mv^{2} =qv'

where v' is the potential difference .

as per the question the charged particle is the helium nucleus whose charge q=3.2*10^{-19} coulomb

the K.E is given as 140 kev i.e 140*10^{3} *1.602*10^{-19} joule

hence the potential difference V'=\frac{K.E}{q}

                                                       =140*10^{3} **1.602*10^{-19} *\frac{1}{3.2*10^{-19} }

                                                      =70.0875*10^{3} volt [ans]

                               




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