Question

A beam of helium-3 atoms (m = 3.016 u) is incident on a target of nitrogen-14 atoms (m = 14.003 u) at rest. During the collision, a proton from the helium-3 nucleus passes to the nitrogen nucleus, so that following the collision there are two atoms: an atom of "heavy hydrogen" (deuterium, m = 2.014 u) and an atom of oxygen-15 (m = 15.003 u). The incident helium atoms are moving at a velocity of 6.346 x 10° m/s. After the collision, the deuterium atoms are observed to be moving forward (in the same direction as the initial helium atoms) with a velocity of 1.531 x 107 m/s.A) What is the final velocity of the oxygen-15 atoms? B) Compare the total kinetic energies before and after the collision.

Answer 1

Answer:

Explanation:

We shall apply law of conservation of momentum to solve the problem .

Helium-3 collides with nitrogen-14 at rest . After the collision the newly formed deuterium atom and oxygen-15  atom moves .

momentum before the collision

= 3.016 x 6.346 x 10⁶ + 14.003 x 0 = 19.14 x 10⁶ unit

momentum after collision

2.014 x 1.531 x 10⁷ + 15.003 V

3.083 x 10⁷ +  15.003 V units

Applying the law of conservation of momentum ,

19.14 x 10⁶ = 3.083 x 10⁷ +  15.003 V

1.914 x 10⁷ = 3.083 x 10⁷ +  15.003 V

15.003 V = - 1.169 x 10⁷

V = .077917 x 10⁷

= 7.79 x 10⁵  m /s

= .0779 x 10⁷ m /s

mass of helium atom = 3.016 u = 3.016 x 1.67 x 10⁻²⁷ kg

velocity = 6.346 x 10⁶ m /s

kinetic energy = 1 /2 x  3.016 x 1.67 x 10⁻²⁷ x (6.346 x 10⁶ )²

= 101.42 x 10⁻¹⁵ J  

kinetic energy of nitrogen atoms = 0

Total energy before collision =  101.42 x 10⁻¹⁵ J  

Similarly kinetic energy after collision

= 1 /2 x [ 2.014 x 1.531² + 15.003 x .0779² ] x 1.67 x 10⁻²⁷ x 10¹⁴

= .835 x [ 4.72  + .09 ] x 10⁻¹³ J

=  4.016 x 10⁻¹³ J

= 401.6 x 10⁻¹⁵  J  

value of kinetic energy is increased .