In 22 we saw Max Planck astound the world of physics by proc

In 2.2 we saw Max Planck astound the world of physics by proclaiming that microscopic oscillators in a black body can have only discrete energies E_n = nhv, where v is the frequency of the oscillator and h = 6.626 Times 10^-34 J-sec. Yet we know that macroscopic oscillators have continuous energies. In this problem we shall explore this apparent contradiction and show that it isn\'t a contradiction at all. Consider the following macroscopic oscillator: two masses m = 1 g are attached to a spring and oscillate with amplitude A = 1.0 cm at frequency v = 100 Hz. Use classical physics to calculate the energy of the oscillator, expressing your result in joules, ergs, and in electron volts (eV). Now, using Planck\'s relation (2.6.1), calculate the corresponding value of n. From your answer to (a), calculate the percentage change in energy that results from unit change in n. Could you measure such a change? The H_2 molecule is a microscopic oscillator. The mass of each hydrogen atom is roughly 1.7 Times 10^-24g. The frequency of oscillation of the two atoms in H_2 is v ~ 1.0 Times 10^14sec^-1. From this data, calculate the energy of this oscillator (in electron volts) for n = 1, 2, and 3. Using classical physics, calculate numerical values (in A) for the amplitude of the three states considered in (c). We see that the amplitude of a microscopic oscillator is quantized.

Solution

a. energy= 1/2 x m x ( 2x pi x 100) ^2 x A^2 = 1/2 x 0.001 x 4 x pi^2 x 10000 x (1 x 10^-2)^2 = 0.01973 J

now hv= 6.626 x 10^-34 x 100 =6.626 x 10^-32 j

n = 0.01973/6.626 x 10^-32= 297766374 x 10^21

b. energy change =0.01 x 0.01973=0.0001973 j

c. same formula of a = 1/2 x 1.7 x 10^-24 x ( 4 x pi^2 x 1x 10^28) x A^2   

A value need to be substituted

6.626 x 10^-34 x 1 x 10^14 =6.626 x 10^20 j

= 6.626 x 10^20 /1.6 x 10^-19= 4.141 x 10^39 eV

so 6.626 x 10^-20 = 1/2 x 1.7 x 10^-27 x ( 4 x pi^2 x 1x 10^28) x A^2

so, A^2 = 0.194 x 10^-21

so, A=1.39 x 10^-11 m