Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei (4He) from gold-197 nuclei (197Au). The energy of the incoming helium nucleus was 7.58 ✕ 10−13 J, and the masses of the helium and gold nuclei were 6.68 ✕ 10−27 kg and 3.29 ✕ 10−25 kg, respectively (note that their mass ratio is 4 to 197. Assume that the helium nucleus travels in the +x-direction before the collision.)
(a)
If a helium nucleus scatters to an angle of 121° during an elastic collision with a gold nucleus, calculate the helium nucleus' final speed (in m/s) and the final velocity (magnitude in m/s and direction counterclockwise from the +x-axis) of the gold nucleus.
(b)
What is the final kinetic energy (in J) of the helium nucleus?
a) Final speed of the helium nucleus: 1.14 x 10^5 m/s
Final velocity of the gold nucleus: 1.14 x 10^5 m/s, 121° counterclockwise from the +x-axis
b) Final kinetic energy of the helium nucleus: 5.76 x 10^-13 J
To solve this problem, we can use the principle of conservation of momentum and conservation of kinetic energy.
(a)
We can start by calculating the final speed of the helium nucleus. We can use the conservation of momentum equation:
m1v1i + m2v2i = m1v1f + m2v2f,
where m1 and v1 are the mass and initial velocity of the helium nucleus respectively, m2 and v2 are the mass and initial velocity of the gold nucleus respectively, and v1f and v2f are the final velocities of the helium and gold nuclei respectively.
Given:
m1 = 6.68 × 10^-27 kg
v1i = 7.58 × 10^-13 J / m1 (using kinetic energy equation KE = 0.5 mv^2)
m2 = 3.29 × 10^-25 kg
v2i = 0 (since the gold nucleus is initially at rest)
v1f = ? (final velocity of helium nucleus)
v2f = ? (final velocity of gold nucleus)
Substituting the given values into the conservation of momentum equation and solving for v1f, we get:
(6.68 × 10^-27 kg) (7.58 × 10^-13 J / 6.68 × 10^-27 kg) + (3.29 × 10^-25 kg) (0) = (6.68 × 10^-27 kg) v1f + (3.29 × 10^-25 kg) v2f.
Simplifying the equation, we find:
5.56 × 10^-14 J + 0 = 6.68 × 10^-27 kg) v1f + 0.
This simplifies to:
v1f = 5.56 × 10^-14 J / 6.68 × 10^-27 kg ≈ 8.32 × 10^12 m/s.
Now, let's calculate the final velocity and direction of the gold nucleus. Since the helium nucleus scatters to an angle of 121°, we can use trigonometry to find the x and y components of the final velocity of the gold nucleus:
v2f_x = -v1f × cos(121°)
v2f_y = -v1f × sin(121°)
Note the negative sign in front of v1f on the right side of the equations because the gold nucleus moves in the opposite direction to the helium nucleus.
Calculating the values,
v2f_x = - (8.32 × 10^12 m/s) × cos(121°) ≈ -3.18 × 10^12 m/s
v2f_y = - (8.32 × 10^12 m/s) × sin(121°) ≈ -7.23 × 10^12 m/s
To find the final velocity of the gold nucleus, we can use the Pythagorean theorem:
v2f = √(v2f_x^2 + v2f_y^2)
Calculating the value,
v2f = √((-3.18 × 10^12 m/s)^2 + (-7.23 × 10^12 m/s)^2) ≈ 7.90 × 10^12 m/s.
So, the final speed of the helium nucleus is approximately 8.32 × 10^12 m/s, and the final velocity (magnitude) and direction of the gold nucleus are approximately 7.90 × 10^12 m/s and counterclockwise from the +x-axis.
(b)
To find the final kinetic energy of the helium nucleus, we can use the equation:
KE = 0.5 mv^2,
where KE is the kinetic energy, m is the mass of the helium nucleus, and v is the final velocity of the helium nucleus.
Given:
m = 6.68 × 10^-27 kg
v = 8.32 × 10^12 m/s (final velocity of helium nucleus)
Plugging the values into the equation, we get:
KE = 0.5 (6.68 × 10^-27 kg) (8.32 × 10^12 m/s)^2 ≈ 2.45 × 10^-13 J.
So, the final kinetic energy of the helium nucleus is approximately 2.45 × 10^-13 J.
To solve this problem, we can apply the conservation of momentum and conservation of energy laws.
(a) Final speed of helium nucleus:
Let the initial velocity of the helium nucleus be v_helium in the +x direction.
After the collision, the helium nucleus scatters to an angle of 121°. Let's call the final speed of the helium nucleus v_helium_f and the final velocity of the gold nucleus v_gold_f.
Using the conservation of momentum:
m_helium * v_helium = m_helium * v_helium_f + m_gold * v_gold_f
Since the helium nucleus was initially moving in the +x direction, its x-component of momentum before and after the collision will be equal:
m_helium * v_helium = m_helium * v_helium_f * cos(121°) + m_gold * v_gold_f * cos(α)
where α is the angle between the final velocity vector of the gold nucleus and the +x-axis.
Using the conservation of energy:
Initial kinetic energy + Initial potential energy = Final kinetic energy + Final potential energy
0.5 * m_helium * v_helium^2 + 0 = 0.5 * m_helium * v_helium_f^2 + 0.5 * m_gold * v_gold_f^2
Substituting the mass ratio of the helium and gold nuclei (4 to 197), and the given values for masses and initial energy into the equations, we can solve for v_helium_f and v_gold_f.
(b) Final kinetic energy of helium nucleus:
The final kinetic energy can be calculated using the equation mentioned above:
Final kinetic energy = 0.5 * m_helium * v_helium_f^2
Substitute the value of v_helium_f obtained from part (a) into this equation to get the final kinetic energy.