The carbon isotope 14C is used for carbon dating of objects. A 14C nucleus can change into a different kind of element, a neighbor on the periodic table with lower mass, by emitting a beta particle – an electron or positron – plus a neutrino or an anti-neutrino. Consider the scenario where 14C ( mass of 2.34 x 10 -26) decays by emitting an electron and anti neutrino. The electron has a mass of 9.11x 10-31 kg and a speed of 4.0 x107 m/s. While the anti neutrino has a momentum of 4.0x10-24 kg-m/s. If the electron and anti neutrino are emitted at right angles from each other, calculate the recoil speed of the nucleus.
The scientific notations should say 2.34 x 10^-26; 9.11 x 10^-31; 4.0 x 10^7; 4.0 x 10^-24; sorry about that.
momentum of electron: mv=9.11E-31 *4.0E7
=3.64400e-23kg-m/s
Momentum of neutrino:4.0x10-24 kg-m/s
these are right triangles
momentumNucleus^2= sume of above squared
= (4.0E-24)^2 + (3.64400e-23)^2 =1.3438736e-45
speed nuceuls= sqrt(above) / massNucleus
= 3.66588816e-23 / 2.34 E-26 = 1567 m/s
check all that
momentum of electron (say in x direction)
Px = 9.11*10^-31 ( 4*10^7 ) = 36.44 * 10^-24
momentum of anti-n (say in y direction)
Py = 4*10^-24
|P| =sqrt( Px^2+Py^2)
= 10^-24 sqrt( 36.44^2 + 16)
= 36.7 * 10^-24
so
m v = 36.7*10^-24
2.34 * 10^-26 v = 36.7 * 10^-24
v = 15.7 * 10^2
v = 1570 m/s
whew, we agree :)
Many thanks!
To solve this problem, we can use the principle of conservation of momentum.
The initial momentum of the system is zero since the initial carbon atom is at rest. Therefore, the total momentum after the decay must also be zero to satisfy the conservation of momentum.
Let's break down the problem into components. We have the momentum in the x-direction (Px) and momentum in the y-direction (Py).
Initially, the momentum in the x-direction is zero, and the momentum in the y-direction is also zero.
After the decay, the electron is emitted in the x-direction, and the anti-neutrino is emitted in the y-direction. The momentum of the electron is given by its mass (m) multiplied by its speed (v) in the x-direction, while the momentum of the anti-neutrino is given directly.
Therefore, the final momentum in the x-direction is m * v, and the final momentum in the y-direction is -4.0x10^(-24) kg-m/s.
Using the Pythagorean theorem, we can find the magnitude of the final momentum (Pf):
Pf = sqrt((Px)^2 + (Py)^2) = sqrt((m * v)^2 + (-4.0x10^(-24))^2)
Now, according to the conservation of momentum, Pf = 0. Therefore, we can set up the equation:
sqrt((m * v)^2 + (-4.0x10^(-24))^2) = 0
Solving the equation:
(m * v)^2 + (-4.0x10^(-24))^2 = 0
Simplifying:
(m * v)^2 = -(-4.0x10^(-24))^2
Taking the square root of both sides:
m * v = sqrt((-4.0x10^(-24))^2)
Calculating the square root:
m * v = 4.0x10^(-24)
Now, solve for the recoil speed (v_final) by dividing both sides by the mass of the carbon nucleus (2.34 x 10^(-26) kg):
v_final = (4.0x10^(-24)) / (2.34 x 10^(-26)) kg-m/s
Calculating the final result will give you the recoil speed of the nucleus.