Question

Alpha decay is nuclear decay in which a helium nucleus is emitted. If the helium nucleus has a mass of 6.80×10^(-27) kg and is given 5.00 MeV of kinetic energy, what is its velocity?

a) 1.65 x 10^7 m/s
b) 2.45 x 10^7 m/s
c) 3.28 x 10^7 m/s
d) 4.91 x 10^7 m/s

Answer 1

The velocity of the helium nucleus emitted during alpha decay is 3.28 x 10^7 m/s.

Step-by-step explanation:

To find the velocity of the helium nucleus, we can use the formula for kinetic energy:

Kinetic Energy = ½mv^2

Where m is the mass and v is the velocity. Rearranging the formula, we have:

v = sqrt(2Kinetic Energy / m)

Substituting the given values: Kinetic Energy = 5.00 MeV = 5.00 x 10^6 eV = 5.00 x 10^6 x 1.6 x 10^-19 J and m = 6.80 x 10^-27 kg, we can calculate the velocity:

v = sqrt(2 x 5.00 x 10^6 x 1.6 x 10^-19 / 6.80 x 10^-27) = 3.28 x 10^7 m/s

Therefore, the correct answer is c) 3.28 x 10^7 m/s.