Question

The value of the angular-frequency variable characteristic of the vibration of an oxygen molecule O2 is 3.58 × 1014 rad/s. What is the difference in energy between the ground state for vibrations and the first excited vibrational state, in electron volts? The mass of a proton is 1.67262 × 10−27 kg , the value of ¯h is 1.05457 × 10−34 J · s, and the atomic mass unit is 1.66054 × 10−27 kg . 1 eV = 1.60218 × 10−19 J Answer in units of eV.

In Niels Bohr’s 1913 model of the hydrogen atom, an electron circles the proton at a distance of 5.42 × 10−11 m with a speed of 1.18 × 106 m/s. The permeability of free space is 1.25664 × 10−6 T · m/A . Compute the magnetic field strength that this motion produces at the location of the proton. Answer in units of T.

Answer 1

The existence of an angular frequency variable w= 3.58 *10^14 rad/s indicates oscillatory motion so use a simple harmonic oscillator as a model.;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; for the O2 molecule, its energy eigenvalues are;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; E(n) = ( n + 0.5) h f ;where w = 2 *pi*f => f = w/( 2*pi );;;;;;;;;;;;;;;;;;;;;;;;;;;;; =>E(n) = ( n + 0.5)h w/(2 *pi) ;;;;;;;;;;;;;;;;;;;;;;;;; ;where n =0 ,1,2,3... that is, for the ground state : E(0) = h w/(4*pi) ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; for the first excited state : E(1) =3hw/(4*pi );;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; so that ....E(1) - E(0) = 3 hw/ ( 4*pi ) - hw/( 4*pi ) = (3-1)hw/(4*pi) = hw/ (2*pi);;;;;;;;;;;;;;;;;;;;;;;;;;;;; = ( 6.63 *10^-34 J-s ) (3.58* 10^14 rad/s )/( 2*pi);;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; = 3.72 * 10^-20 J

=(3.72*10 ^-20 J)/(1.6 * 10^-19 J/eV ) = 0.2325 eV