Question

A 2kg block oscillates with an amplitude of 10cm on a spring of force constant 120 N/m .

(a) In which quantum state is the block?

(b) The block has a slight electric charge and drops to a lower energy level by generating a photon. What is the minimum energy decrease possible, and what would be the corresponding fractional change in energy?

Short answer

(a) The block is in the 7.37 x 10³² number quantum state.

(b) The minimum energy decrease possible is 8.14 x 10^{-34 J} . The corresponding fraction change in energy is 13.57 x 10^-34.

Step-by-step solution

Given data

Mass of the block is

m = 2 kg

Amplitude of oscillation is

$\begin{array}{rcl}a& =& 10\text{cm}\\ & =& 10·\left(1\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ & =& 0.1\text{m}\end{array}$

Spring constant is

k = 120 N/m

Total energy of oscillation, frequency of oscillation and energy of quantum harmonic oscillator

Total energy of oscillation is

$E=\frac{1}{2}k{a}^2$ .....(I)

Frequency of oscillation is

$\omega =\sqrt{\frac{k}{m}}$ .....(II)

Energy of a quantum harmonic oscillator is

$E_n=\left(n+\frac{1}{2}\right)\hslash \omega$ ,,,,,(III)

Here $\hslash$is the reduced Planck's constant of value

$\hslash =1.05\times {10}^{-34}J·s$

and n is the quantum state.

Determining the quantum state

From equation (I), the energy of the oscillator is

$\begin{array}{rcl}E& =& \frac{1}{2}\times 120\text{N/m}\times {\left(0.1\text{m}\right)}^2\\ & =& 0.6·\left(1\text{N}·\text{m}\times \frac{1\text{J}}{1\text{N}·\text{m}}\right)\\ & =& 0.6\text{J}\end{array}$

From equation (II) the frequency of oscillation is

$\begin{array}{rcl}\omega & =& \sqrt{\frac{120\text{N/m}}{2\text{kg}}}\\ & =& 7.75·\sqrt{\left(1\text{N}\times \frac{1\text{kg}·{\text{m/s}}^2}{1\text{N}}\right)·\left(\frac{1}{1\text{m}}\right)·\left(\frac{1}{1\text{kg}}\right)}\\ & =& 7.75{\text{s}}^{-1}\\ & & \end{array}$

Thus from equation (III),

$\begin{array}{rcl}{E}_n& =& E\\ \left(n+\frac{1}{2}\right)\hslash \omega & =& 0.6\text{J}\\ n& =& \frac{0.6\text{J}}{1.05\times {10}^{-34}\text{J}·\text{s}\times 7.75{\text{s}}^{-1}}-\frac{1}{2}\\ & \approx & 7.37\times {10}^{32}\\ & & \end{array}$

Thus the quantum state is $7.37\times {10}^{32}$.

Determining the minimum change in energy

From equation (III) the minimum change in energy is

$\begin{array}{rcl}\Delta E& =& \hslash \omega \\ & =& 1.05\times {10}^{-34}\text{J}·\text{s}\times 7.75{\text{s}}^{-1}\\ & =& 8.14\times {10}^{-34}\text{J}\end{array}$

The corresponding fractional change in energy is

$\begin{array}{rcl}\frac{\Delta E}{E}& =& \frac{8.14\times {10}^{-34}\text{J}}{0.6\text{J}}\\ & =& 13.57\times {10}^{-34}\\ & & \end{array}$

Thus the fractional change in energy is $13.57\times {10}^{-34}$.