27/13AL+? ->> 24/11NA+4/2HE

 ¹₀n

Explanation:

The given expression is:

          ²⁷₁₃Al      +     ₐᵇT   →   ²⁴₁₁Na    +    ⁴₂He

This expression is typical of nuclear reactions. To balance this expression we must conserved the mass number and the atomic number.

The mass number is the superscripts and the atomic number is the subscript;

 Now;

 conserving mass number:

             27 + b  = 24 + 4

             27 + b  = 28  

                   b  = 28 - 27  = 1

conserving the atomic number;

            13 + a  = 11 + 2

             13 + a  = 13

                    a  = 0

So;

The particle T is a neutron    ¹₀n

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally