A molecule has a rotational inertia 14000amupmsquare spining at an angular speed of 4. 30x10exponent12 radpersecond
a) express the rotational inertia in kgmsq
b)calculate rotational kinetic energy in EV.
To convert rotational inertia from atomic mass units times picometer squared (amu pm^2) to kilogram meter squared (kg m^2), we need to use the following conversion factors:
1 amu = 1.66053886 x 10^-27 kg
1 pm = 1 x 10^-12 m
a) To convert rotational inertia from amu pm^2 to kg m^2, we multiply by the appropriate conversion factors:
Rotational inertia (kg m^2) = (Rotational inertia in amu pm^2) * (1.66053886 x 10^-27 kg / amu) * (1 x 10^-12 m / pm)
Rotational inertia (kg m^2) = 14000 amu pm^2 * (1.66053886 x 10^-27 kg / amu) * (1 x 10^-12 m / pm)
Simplifying the units gives:
Rotational inertia (kg m^2) = 14000 * 1.66053886 x 10^-39 kg m^2
Rotational inertia (kg m^2) ≈ 2.3247532 x 10^-34 kg m^2
Therefore, the rotational inertia in kg m^2 is approximately 2.3247532 x 10^-34 kg m^2.
b) To calculate the rotational kinetic energy in electron volts (eV), we use the following conversion factor:
1 J = 6.242 x 10^18 eV
The formula for rotational kinetic energy is:
Rotational kinetic energy (J) = (1/2) * rotational inertia (kg m^2) * (angular speed (rad/s))^2
Rotational kinetic energy (J) = (1/2) * 2.3247532 x 10^-34 kg m^2 * (4.30 x 10^12 rad/s)^2
Rotational kinetic energy (J) ≈ 8.76394 x 10^-22 J
To convert the rotational kinetic energy to electron volts (eV), we use the conversion factor:
Rotational kinetic energy (eV) = (8.76394 x 10^-22 J) * (6.242 x 10^18 eV / J)
Rotational kinetic energy (eV) ≈ 5.46472 x 10^-3 eV
Therefore, the rotational kinetic energy is approximately 5.46472 x 10^-3 eV.
a) To express rotational inertia in kg·m², we need to convert the given rotational inertia from atomic mass units (amu)·pm² (picometers squared) to kg·m².
To convert from amu·pm² to kg·m², we need the following conversion factors:
1 amu = 1.66054 x 10^-27 kg
1 pm = 1 x 10^-12 m
Therefore, to convert from amu·pm² to kg·m², we need to use the following conversion factors:
(1 amu·pm²) x (1.66054 x 10^-27 kg / 1 amu) x (1 x 10^-12 m / 1 pm) = 1.66054 x 10^-39 kg·m²
Now, let's calculate the rotational inertia in kg·m²:
Rotational inertia (kg·m²) = 14000 amu·pm² x (1.66054 x 10^-39 kg·m² / 1 amu·pm²)
Rotational inertia (kg·m²) ≈ 2.324756 x 10^-34 kg·m²
Therefore, the rotational inertia of the molecule is approximately 2.324756 x 10^-34 kg·m².
b) To calculate the rotational kinetic energy in electron volts (eV), we need to use the following formula:
Rotational kinetic energy (eV) = (1/2) * I * ω²
Where:
I = rotational inertia (kg·m²)
ω = angular speed (rad/s)
So, let's substitute the given values into the formula to calculate the rotational kinetic energy:
Rotational kinetic energy (eV) = (1/2) * (2.324756 x 10^-34 kg·m²) * (4.30 x 10^12 rad/s)²
Now, let's perform the calculation:
Rotational kinetic energy (eV) ≈ 2.324756 x 10^-34 kg·m² * (1/2) * 4.30² x 10^24 eV
Rotational kinetic energy (eV) ≈ 2.324756 x 10^-34 kg·m² * 9.25 x 10^24 eV
Rotational kinetic energy (eV) ≈ 2.150814 x 10^-9 eV
Therefore, the rotational kinetic energy of the molecule is approximately 2.150814 x 10^-9 electron volts (eV).