Calculate the kinetic energy of `O_(2)` molecule which has wavelength of 2.5 pm.

Answer

`6.61xx10^(-19)J`

Explanation

Step I. Calculation of velocity of oxygen molecule :
According to de Broglie equaition,
`lambda=(h)/(mv) or v=(h)/(m lambda) , h=6.626xx10^(-34) "kg" m^(2)s^(-1)`,
`lambda=2.5xx10^(-12)m , m_(O_(2))=(0.032)/(6.022xx10^(23)) `kg
`v=((6.626xx10^(-34) "kg"m^(2)s^(-1))xx(6.022xx10^(23)))/((0.032 "kg")xx(2.5xxx10^(-12)m))=4987.7m s^(-1)`.
Step II. Calculation of kinetic energy of oxygen molecule :
`K.E.=1//2mv^(2)=(1)/(2)xx((0.032)/(6.022xx10^(23))kg)xx(4987.7ms^(-1))^(2)`
`=6.61xx10^(-19) kg m^(2)s^(-2)=6.61xx10^(-19) J`.