An alpha particle (which is a helium nucleus, Q = +2e,\; m = 6.64 x 10^{-27} kg ) is emitted in a radioactive decay with KE = 5.53 MeV. What is its speed?

This is Giancoli Answers with Mr. Dychko. Kinetic energy is one-half mass times velocity squared. And you can multiply both sides by two and divide both sides by m and then take the square root of both sides, and then you can get that the speed is the square root of two times kinetic energy divided by mass. The kinetic energy of this alpha particle is 5.53 times 10 to the six electron volts. The megaelectron volts means times 10 to the six electron volts. And multiply that by 1.6 times 10 to the minus 19 Joules per electron volts so that you're left with units with Joules on top. And divided by the mass of the alpha particle which is 6.67 times 10 to the negative 27 kilograms, and you get 1.63 times 10 the seven meters per second. It is its speed.

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An alpha particle (which is a helium nucleus, Q=+2e, m=6.64×10−27 kgQ = +2e,\; m = 6.64 \times 10^{-27} \textrm{ kg}Q=+2e,m=6.64×10−27 kg) is emitted in a radioactive decay with KE = 5.53 MeV. What is its speed?

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An alpha particle (which is a helium nucleus, $Q = +2e,\; m = 6.64 \times 10^{-27} \textrm{ kg}$ ) is emitted in a radioactive decay with KE = 5.53 MeV. What is its speed?