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**.

\[Initial \: momentum = Final\: momentum\]

**Expert Answer**

**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}$

\[v_{1} = \dfrac{mv – m_{2}v_{2} }{m_{1}} \]

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

\[v_{1} = 3533 \dfrac{m}{s}\]