Question

Beta decay is nuclear decay in which an electron is emitted. If the electron is given 0.750 MeV of kinetic energy, what is its velocity?

a) 0.985c
b) 0.750c
c) 0.600c
d) 1.200c

Answer 1

The velocity of the electron emitted in beta decay, given 0.750 MeV of kinetic energy, is approximately 0.985 times the speed of light (c).

Step-by-step explanation:

When a beta decay occurs, an electron is emitted from the nucleus. To determine the velocity of the electron, we can use the equation:

KE = (1/2)mv^2

where KE is the kinetic energy, m is the mass of the electron, and v is the velocity of the electron. Given that the electron is given 0.750 MeV of kinetic energy, we can convert this to joules using the conversion factor 1 MeV = 1.6 x 10^-13 J. Plugging in the values, we can solve for v:

0.750 MeV = (1/2)(9.11 x 10^-31 kg)v^2

v = √(2(0.750 MeV) / (9.11 x 10^-31 kg))

v ≈ 0.985c

Therefore, the velocity of the electron is approximately 0.985 times the speed of light (c).