Question

A helium nucleus (charge +2e) speeds by moving across a potential difference of -5kV. It starts...

A helium nucleus (charge +2e) speeds by moving across a potential difference of -5kV. It starts from rest. How fast is it moving after it crosses the potential difference? Remember a helium nucleus has a mass of 34 atomic units (u) and 1u=1.66*10^-27kg.

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Answer #1

Using energy conservation on helium nucleus

dKE = dPE

dKE = KEf - KEi

KEi = 0 J, since initial speed is zero

dKE = KEf = 0.5*m*Vf^2

dPE = Change in potential energy = q*dV

q = charge on helium = 2e

dV = -5 kV

So,

KEf = 0.5*m*Vf^2 = 2*e*dV

SEE THAT MASS OF HELIUM NUCLEUS SHOULD BE 4 ATOMIC UNITS, PLEASE CHECK YOUR VALUE AGAIN AND LET ME KNOW IF YOU STILL WANT ME SOLVE IT WITH 4 ATOMIC UNITS.

Vf = sqrt (4*e*dV/m)

Vf = sqrt (4*1.6*10^-19*5*10^3/(4*1.66*10^-27))

Vf = 6.94*10^5 m/sec

Please Upvote.

Again I will solve it with your given mass of 34 atomic unit if this one doesn't work. Let me know in the comments, but mass of helium should be 4 atomic unit. So please before downvote let me know in the comment section.

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