Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chem
ID: 1790736 • Letter: E
Question
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 7.50 1013 J, and the masses of the helium and gold nuclei were 6.68 1027 kg and 3.29 1025 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 141° 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.
4He speed ??? m/s 197
Au velocity ???m/s 197
Au direction ???° counterclockwise from the +x-axis
(b)What is the final kinetic energy (in J) of the helium nucleus?
Explanation / Answer
Let's agree to put the origin of the coordinate system at the site of the stationary gold nucleus and the +x direction in the direction in which the incoming helium nucleus travels. A statement of conservation of energy is written as Equation (1).
mHevi He2/2 = mHevf He2/2 + mAuvf Au2/2 .................(1)
A statement of conservation of momentum in the x direction may be written as Equation (2).
mHevi He = mHevf He cos + mAuvf Au cos ................... (2)
mHevf He sin = mAuvf Au sin ................(3)
At this point, we have three equations ((1), (2), and (3)) and three unknowns
(vf He, vf Au, and ).
We should be able to solve these three equations simultaneously in order to obtain the three unknowns. While the associated algebra may be done in a number of ways, the following is one of the more straightforward.
Let's first attempt to eliminate the angle . Looking at Equations (2) and (3), we see that if we can manipulate these two equations in such a manner as to obtain
cos2 + sin2,
we will be able to eliminate since
cos2 + sin2 = 1.
Square Equation (3) to obtain (3a).
mAu2vf Au2 sin2 = mHe2vf He2 sin2…… (3a)
Rearrange (2) in order to isolate the cos term and then square this equation to get
mAu2vf Au2 cos2 = mHe2vi He2 2mHe2vi Hevf He cos + mHe2vf He2 cos2………(4)
Adding Equations (3a) and (4), we obtain
mAu2vf Au2(cos2 + sin2) = mHe2vf He2(sin2 + cos2) + mHe2vi He2 2mHe2vi Hevf He cos .……….(5)
Recognizing that
(cos2 + sin2 = 1) and (sin2 + cos2 = 1),
we may rewrite (5) as (5a).
mAu2vf Au2 = mHe2vf He2 + mHe2vi He2 2mHe2vi Hevf He cos ………..(5a)
Next notice that we have the term mAu2vf Au2
in Equation (5a) and in Equation (1), after we cancel the 2 and multiply each term by mAu. Equating these, we have Equation (6),
mAu2vf Au2 = mHe2(vf He2 + vi He2 2vi Hevf He cos ) = mHemAu(vi He2 vf He2),…….(6)
which may be rewritten as (6a).
vf He2(1+ (mAu/mHe))
+ vi He2(1 - (mAu/mHe))
2vi Hevf He cos = 0………….(6a)
Inserting the ratio mAu/mHe
and the value for the angle , we have the quadratic
vf He2(1+ (197/4))
+ vi He2(1 - (197/4))
2vi Hevf He cos141 = 0………. (7)
or
(50.3)vf He2 + 1.55*vi Hevf He (48.3)vi He2 = 0…………….(7a)
We can determine vi He from the information that
Ki He = mHevi He2/2
Or
vi He2 = 2KiHe/mHe = (2*7.5*10^-13)/(6.68*10^-27) = 2.25*10^14
vi He = 1.5*10^7 m/s
Inserting vi He2 and vi He into (7a), we have the following quadratic,
(50.3)vf He2 + 1.55*1.5*10^7*vf He (48.3)(2.25*10^14) = 0,…………(8)
which may be solved for vf He to obtain
vf He = 1.45*10^7 m/s
Using the basic definition of kinetic energy, we may rewrite the energy statement (Equation (1)) as
2Ki He = mHevf He2 + mAuvf Au2…………(9)
Inserting values, we have
2*7.5*10^-13 = 6.68*10^-27*(1.45*10^7)^2 + 3.29*10^-25*vf Au2………(9a)
which may be solved for vf Au to obtain
vf Au = 5.4*10^5 m/s
Finally, Equation (3) may be solved for sin and values inserted to obtain
sin = (mHevf He sin )/(mAuvf Au) = (mHe/mAu)*(vf He/vf Au)* sin
Inserting values, we obtain
sin = (4/197)*(145*10^5/5.4*10^5)*sin141 = 0.343 ===========> = 20.1°
In summary, the final speed of the helium nucleus is vf He = 1.45*10^7 m/s
and the final velocity (magnitude and direction) of the gold nucleus is vf Au = 5.4*10^5 m/s
directed at an angle of
= 20.1°.
mHevi He2/2 = mHevf He2/2 + mAuvf Au2/2 .................(1)
A statement of conservation of momentum in the x direction may be written as Equation (2).
mHevi He = mHevf He cos + mAuvf Au cos ................... (2)
A statement of conservation of momentum in the y direction may be written as Equation (3).mHevf He sin = mAuvf Au sin ................(3)
At this point, we have three equations ((1), (2), and (3)) and three unknowns
(vf He, vf Au, and ).
We should be able to solve these three equations simultaneously in order to obtain the three unknowns. While the associated algebra may be done in a number of ways, the following is one of the more straightforward.
Let's first attempt to eliminate the angle . Looking at Equations (2) and (3), we see that if we can manipulate these two equations in such a manner as to obtain
cos2 + sin2,
we will be able to eliminate since
cos2 + sin2 = 1.
Square Equation (3) to obtain (3a).
mAu2vf Au2 sin2 = mHe2vf He2 sin2…… (3a)
Rearrange (2) in order to isolate the cos term and then square this equation to get
mAu2vf Au2 cos2 = mHe2vi He2 2mHe2vi Hevf He cos + mHe2vf He2 cos2………(4)
Adding Equations (3a) and (4), we obtain
mAu2vf Au2(cos2 + sin2) = mHe2vf He2(sin2 + cos2) + mHe2vi He2 2mHe2vi Hevf He cos .……….(5)
Recognizing that
(cos2 + sin2 = 1) and (sin2 + cos2 = 1),
we may rewrite (5) as (5a).
mAu2vf Au2 = mHe2vf He2 + mHe2vi He2 2mHe2vi Hevf He cos ………..(5a)
Next notice that we have the term mAu2vf Au2
in Equation (5a) and in Equation (1), after we cancel the 2 and multiply each term by mAu. Equating these, we have Equation (6),
mAu2vf Au2 = mHe2(vf He2 + vi He2 2vi Hevf He cos ) = mHemAu(vi He2 vf He2),…….(6)
which may be rewritten as (6a).
vf He2(1+ (mAu/mHe))
+ vi He2(1 - (mAu/mHe))
2vi Hevf He cos = 0………….(6a)
Inserting the ratio mAu/mHe
and the value for the angle , we have the quadratic
vf He2(1+ (197/4))
+ vi He2(1 - (197/4))
2vi Hevf He cos141 = 0………. (7)
or
(50.3)vf He2 + 1.55*vi Hevf He (48.3)vi He2 = 0…………….(7a)
We can determine vi He from the information that
Ki He = mHevi He2/2
Or
vi He2 = 2KiHe/mHe = (2*7.5*10^-13)/(6.68*10^-27) = 2.25*10^14
vi He = 1.5*10^7 m/s
Inserting vi He2 and vi He into (7a), we have the following quadratic,
(50.3)vf He2 + 1.55*1.5*10^7*vf He (48.3)(2.25*10^14) = 0,…………(8)
which may be solved for vf He to obtain
vf He = 1.45*10^7 m/s
Using the basic definition of kinetic energy, we may rewrite the energy statement (Equation (1)) as
2Ki He = mHevf He2 + mAuvf Au2…………(9)
Inserting values, we have
2*7.5*10^-13 = 6.68*10^-27*(1.45*10^7)^2 + 3.29*10^-25*vf Au2………(9a)
which may be solved for vf Au to obtain
vf Au = 5.4*10^5 m/s
Finally, Equation (3) may be solved for sin and values inserted to obtain
sin = (mHevf He sin )/(mAuvf Au) = (mHe/mAu)*(vf He/vf Au)* sin
Inserting values, we obtain
sin = (4/197)*(145*10^5/5.4*10^5)*sin141 = 0.343 ===========> = 20.1°
In summary, the final speed of the helium nucleus is vf He = 1.45*10^7 m/s
and the final velocity (magnitude and direction) of the gold nucleus is vf Au = 5.4*10^5 m/s
directed at an angle of
= 20.1°.
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