 # An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the remaining nucleus slows to 350 m/s. If the alha particle has a mass of 4.0u and the original nucleus has a mass of 222u. What speed does the alpha particle have when it is emitted? This article aims to find the speed of the alpha particle after it is emitted. The article uses the principle of conservation of linear momentum. The principle of conservation of momentum states that if two objects collide, then total momentum before and after collision will be same if there is no external force acting on colliding objects.

Conservation of linear momentum formula mathematically expresses that momentum of the system remains constant when the net external force is zero.

$Initial \: momentum = Final\: momentum$

Given

The mass of the given nucleus is,

$m = 222u$

The mass of the alpha particle is,

$m_{1} = 4u$

The mass of the new nucleus is,

$m_{2} = (m – m_{ 1 })$

$= (222u – 4u ) =218u$

The speed of the atomic nucleus before emission is,

$v = 420 \dfrac{m}{s}$

The speed of the atomic nucleus after emission is,

$v = 350 \dfrac{m}{s}$

Let’s suppose the speed of the alpha is $v_{1}$. Using the principle of conservation of linear momentum we have,

$mv = m _ { 1 } v _ { 1 } + m _ { 2 } v _{ 2 }$

Solve the equation for unknown $v_{1}$

$v _ { 1 } = \dfrac { m v – m _ { 2} v _ { 2 } } { m_ { 1} }$

$= \dfrac { ( 222u ) ( 420 \dfrac { m }{s }) – ( 218 u ) ( 350 \dfrac { m } { s } ) } { 4 u }$

$v _ { 1 } = 4235 \dfrac { m } { s }$

## Numerical Result

The speed of the alpha particle when it is emitted is $4235 m/s$.

## Example

An atomic nucleus initially moving at $400 m/s$ emits an alpha particle in the direction of its velocity and remaining nucleus slows down to $300 m/s$. If an alpha particle has a mass of $6.0u$ and the original nucleus has a mass of $200u$, what is the speed of an alpha particle when it is emitted?

Solution:

The mass of the given nucleus is,

$m = 200u$

The mass of the alpha particle is,

$m_{1} = 6u$

The mass of the new nucleus is,

$m _ { 2 } = ( m – m _ { 1 } )$

$= ( 200 u – 6 u ) = 194 u$

The speed of the atomic nucleus before emission is,

$v = 400 \dfrac { m } { s }$

The speed of the atomic nucleus after emission is,

$v = 300 \dfrac{m}{s}$

Let’s suppose the speed of the alpha is $v_{1}$. Using the principle of conservation of linear momentum, we have,

$mv = m _ { 1 } v_{1} + m_{2} v_{2}$

Solve the equation for unknown $v_{1}$

$v_{1} = \dfrac{mv – m_{2}v_{2} }{m_{1}}$

$= \dfrac{( 200u)(400\dfrac{m}{s}) – ( 196u )( 300\dfrac{m}{s})}{6u}$

$v_{1} = 3533 \dfrac{m}{s}$