Consider translational motion of single O2 molecule (mass = 5.314 * 10^-26 kg) t
ID: 1035617 • Letter: C
Question
Consider translational motion of single O2 molecule (mass = 5.314 * 10^-26 kg) trapped in one-dimensional box potential ("particle in a box), which has a width of 5 cm.
A) What is the energy difference between the two lowest quantized energy levels?
B) The amount of classical translational thermal energy for a molecule confined in one dimension is 1/2kT where T is the temperature and k is the Boltzmann constant (1.381 * 10^-23 J/K). If the temperature is 300 K, at what value of n would the quantized O2 energy exceen the thermal energy?
C) Using the vaue of n from b), calculate the energy seperation between the adjacent levels (ie, n vs n+1)
Explanation / Answer
A) En=n^2 h^2/8mL^2
where n=energy level,
h=planck's constant=6.626*10^-34 Js
m=mass of particle=5.314 * 10^-26 kg
L=0.05m
for n=1 (lowest level),n=2
En=2 -En=1=3h^2/8mL^2=3*(6.626*10^-34 Js)^2/(8*(5.314 * 10^-26 kg)(0.05m)^2)=1239.288*10^-42 J=1.239*10^-39J
energy difference between the two lowest quantized energy levels=1.239*10^-39J
B) Translational thermal energy=1/2*(1.381 * 10^-23 J/K)*300K=2.071*10^-21 J
En=n^2 h^2/8mL^2=n^2 (6.626*10^-34 Js)^2/(8*(5.314 * 10^-26 kg)(0.05m)^2)=0.416*10^-39J
En=n^2*(0.416*10^-39J)=1.239*10^-39J
So,n^2=(1.239*10^-39J)/(0.416*10^-39J)=2.978
n=1.72=2(rounded off to nearest integer)
n=2
C) n=2 ,n=n+1=2+1=3
E3-E2=(3^2- 2^2)*h^2/8mL^2=5*(0.416*10^-39J)=2.08*10^-39J