Physics, 21.04.2020 01:19 abby186

Ernest Rutherford (who was the first person born in New Zealand 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). During this experiment, the energy of the incoming helium nucleus was 8.00 × 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).

a. If a helium nucleus scatters to an angle of 1200 during an elastic collision with a gold nucleus, calculate the helium nucleus's final speed and the final velocity (magnitude and direction) of the gold nucleus.
b. What is the final kinetic energy of the helium nucleus?

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Answer from: Shybaby7334

the velocity of the gold nuclei $v'_2 = 5.34 *10^5 \ m/s$

the angle of the velocity of gold nuclei = 29.64° clockwise (i.e down the horizontal)

the final kinetic energy of helium nucleus is $7.52 *10^{-13} \ \ J$

Explanation:

The formula for kinetic Energy (K.E) is:

$K.E = \frac{1}{2}m_1v_1^2$

$v_1 = (\frac{2K.E}{m_1})^{1/2}$

where;

m_1 = mass of the helium

v_1 = velocity of helium

replacing $8.00*10^{-13} \ J \ for \ K.E$ and $m \ with \ 6.68*10^{-27} \ kg$ ; we have :

$v_1 = (\frac{2*8.00*10^{-13 \ }J}{6.68*10^{-27} kg})^{1/2}$

$v_1 = 1.548*10^7 \ m/s$

Applying conservation of momentum along x- axis since the collision is elastic

$m_1v_1 =m_1v'_1cos \theta _1 + m_2v_2' cos \theta_2$ ------ equation (1)

where

$m_1 \ and \ m_2$ = mass of helium and gold respectively

v_1 = initial velocity of helium

$v'_1 \ and \ v'_2$ = final velocity of gold

$\theta _1 \ and \ \theta _2$ = scattered angle for helium and gold respectively

Along the y - axis; the equation for conservation of momentum is :

$0 = m_1 v_1' \theta_1 + m_2 v_2' sin \theta _2$ ---- equation (2)

Equating equation (1) and (2); we have:

$(v'_2)^2 = \frac{m_1^2}{m_2^2}[(v_1)^2 - 2 v_1v_1' xos \theta _1 + (v'_1)^2]$ ----- equation (3)

However, the conservation of internal kinetic energy guves:

$\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_1(v_1')^2 + \frac{1}{2}m_2(v'_2)^2 \frac{m_1}{m_2}$

Making (v_2')^2 the subject of the formula ; we have:

$(v_2')^2 = \frac{m_1}{m_2}(v_1^2-(v_1')^2)$ ----- equation (4)

Replacing the expression of (v'_2)^2 in equation (3) into equation (4) ; we have

$[(1+\frac{m_1}{m_2})(v_1')^2-2(\fracm_1}{m_2}(v_1'cos \theta _1)v_1 - (1-\frac{m_1}{m_2})(v_1)^2] = 0$

In the above expression;

replacing ;

$1.548*10^7 \ m/s$ $for \ v_1$ ; $\theta = 120^0$ ; $m_1 = 6.68*10^{-27} \ kg$ ; $m_2 = 3.29*10^{-25}\ kg$ ; we have:

$[1 + ( \frac{6.68*10^{-27}}{3.29*10^{-25}})(v_1')^2 - ( 2(\frac{6.68*10^{-27}}{3.29*10^{-25}}) (1.548*10^7)cos 120^0)v'_1-(1-\frac{6.68*10^{-27}}{3.29*10^{-25}})(1.548*10^7)^2 = 0$

$1.02v^2 - (3.143 *10^5)v -2.348*10^{14} = 0$

The above is a quadratic equation; now solving by using the quadratic formula; we have:

$v_1' = \frac{-3.143*10^5 \pm \sqrt{(3.143*10^5)^2 -4(1.02)(-2.348*10^{14})}}{2(1.02)}$

since we are considering the positive value from the above expression; we have

$v'_1 = 1.50*10^7 \ m/s$

NOW; we substitute our known values into equation (4) in order to solve for v_2' ; we have:

$(v'_2 )^2 = \frac {6.68*10^{23}}{3.29*10^{-23}}((1.548*10^7)^2-(1.502*10^7)^2)$

$(v'_2 )^2 =2.848*10^{11} \ m^2/s^2$

$(v'_2 )=\sqrt{2.848*10^{11} \ m^2/s^2}$

$v'_2 = 5.34 *10^5 \ m/s$

Therefore; the velocity of the gold nuclei $v'_2 = 5.34 *10^5 \ m/s$

From equation (2)

$0 = m_1 v_1' \theta_1 + m_2 v_2' sin \theta _2$

Therefore replacing our known values and solving for $\theta_2$ ; we have:

$sin \theta _2 =\frac{(6.68*10^{-27})(1.50*10^7)sin 120^0}{(3.29*10^{-25}(5.34*10^5)}$

$sin \theta _2 = -0.4946$

$\theta _2 =sin^{-1}( -0.4946)$

$\theta _2 = -29.64^0$

∴ the angle of the velocity of gold nuclei = 29.64° clockwise (i.e down the horizontal)

b) Equation to determine the final kinetic energy (K.E_f) of helium is:

$K.E_f = \frac{1}{2}m_1(v_1')^2$

$= \frac{1}{2}(6.68*10^{-27})}(1.50*10^7)^2$

= $7.52 *10^{-13} \ \ J$

Thus, the final kinetic energy of helium nucleus is $7.52 *10^{-13} \ \ J$

I hope this explanation helps alot!.

Answer from: Quest

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006 kilo calorie

then ,802.3 kilo joule will be = 802.3 $\times$ 0.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

Answer from: Quest

Answer8.0m;

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Ernest Rutherford (who was the first person born in New Zealand to be awarded the Nobel Prize in che...