Question

A bare helium nucleus has two positive charges and a mass of 6.64 ✕ 10-27 kg. (a) Calculate its kinetic energy in joules at 4.60% of the speed of light. J. (b) What is this in electron volts? eV. (c) What voltage would be needed to obtain this energy? V

Answer 1

Explanation:

Mass of the positive charge = $6.64\times 10^{-27} kg$

Velocity of the positive charge :4.60% of speed of light =$3\times 10^8 m/s\times \frac{4.60}{100}=1.38\times 10^7 m/s$

$K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 6.64\times 10^{-27} kg\times (1.38\times 10^7 m/s)^2=6.322\times 10^{-13} J$

a) Kinetic energy of two positive charges =$2\times 6.322\times 10^{-13} J=1.264\times 10^{-12} J$

b) 1 Joule = $6.642\times 10^{18} eV$

Energy in electron volts =$1.264\times 10^{-12} J\times 6.642\times 10^{18} eV=8.39\times 10^{6} eV$

c) $K.E=qV$ here two positive charges in helium nucleus

$1.264\times 10^{-12} J=2\times 1.602\times 10^{-19} C\times V$

$V=\frac{1.264\times 10^{-12} J}{2\times 1.602\times 10^{-19} C}=3940000 V=3940 kiloVolts$

Voltage required will be of 3940 kiloVolts

Answer 2

The relativistic mass is almost identical to the rest mass.
If the rest mass is exactly 6.64000 x 10⁻²⁷ kg, then the relativistic
mass, rounded to the third significant figure, is  6.65 x 10⁻²⁷ kg.

a).  Kinetic energy = (1/2) (mass) (speed)²

Speed = (0.046) x (299,792,458 m/s)

         KE = (1/2) (6.65 x 10⁻²⁷) (1.902 x 10¹⁴)  =  6.32 x 10⁻¹³ joule


b).  (6.32 x 10⁻¹³ joule) x (6.242 x 10¹⁸ eV/joule)  =  3.947 x 10⁶  eV


c).  1 elementary charge = 1.60 x 10⁻¹⁹ Coulomb

      Charge of 1 Helium nucleus = 2 elementary charges = 3.20 x 10⁻¹⁹ C.

 6.32 x 10⁻¹³ joule / 3.20 x 10⁻¹⁹ C  =  1,975,000 joule/Coulomb

                                                          =   1,975 kV .

That's an awful lotta volts for a straight accelerator.
I see two possibilities:

a).  I made a mistake in my orders of magnitude.  If so, I leave it
to you to find my mistake.  I feel that I've already contributed enough
to earn 5 points. 

b).  If the math is true, then in order to accelerate alpha particles to
3.95 MeV, you'll need some sort of a pulsed scheme, as in a cyclotron,
synchrotron, or the LHC if you can buy some time.


You're not too generous with your points, are you Pap3, or whatever
your name is today, bless your shiny roboticles.