Question

An atom of beryllium (m8.00 u) splits into two atoms of helium (m4.00 u) with the release of 92.2 keV of energy. If the original beryllium atom is at rest, find the kinetic energies and speeds of the two helium atoms. kinetic energy 6.1 speed kev X m/s

Answer 1

The kinetic energy and speed of the two helium atoms are
`73.6*10^(-16)\ J` and
`14.88*10^(5)\ m/s`.

Step-by-step explanation:

Given that,

Mass of beryllium = 8.00 u

Mass of helium = 4.00 u

Energy = 92.2 keV

We need to calculate the kinetic energy

The kinetic energy of the helium atom is the half of the total kinetic energy released.

`K.E_(h)=(E_(total))/(2)`

`K.E_(h)=(92.0)/(2)`

`K.E_(h)=46 K.eV`

Therefore, the kinetic energy of the each helium atom is

`K.E_(h)=(1)/(2)mv^2`

`K.E_(h)=(92.0*10^3*1.6*10^(-19))/(2)`

`K.E_(h)=73.6*10^(-16)\ J`

We need to calculate the speed

Using formula of speed

`(1)/(2)* mv^2=73.6*10^(-16)\ J`

`v^2=2*(73.6*10^(-16))/(m)`

`v^2=2*(73.6*10^(-16))/(4.00*1.66*10^(-27))`

`v^2=221.68*10^(10)`

`v=14.88*10^(5)\ m/s`

Hence, The kinetic energy and speed of the two helium atoms are
`73.6*10^(-16)\ J` and
`14.88*10^(5)\ m/s`.