Question

Question 1: Assume that the pendulum of a grandfather clock acts as one of Planck'sresonators. If it carries away an energy of 8.1 x 10-15 eV in a one-quantumchange, what is the frequency of the pendulum? (Note that an energy this smallwould not be measurable. For this reason, we do not notice quantum effects in thelarge-scale world.)

Answer 1

Given:

Energy = 8.1 x 10⁻¹⁵ eV.

Let's find the frequency of the pendulum.

To find the frequency, apply the formula for the energy of a light quantum:

`E=hf`

Where:

E is the energy

h is Planck's constant = 6.63 x 10⁻³⁴ m² kg/s

f is the frequency.

Where:

1 eV = 1.6 x 10⁻¹⁹ J.

Rewrite the formula for f and solve:

`f=(E)/(h)`

Thus, we have:

`f=(8.1*10^(-15)*(1.6*10^(-19)))/(6.63*10^(-34))`

Solving further:

`\begin{gathered} f=\frac{8.1*10^(-15)*(1.6*10^(-19))}{6.63*10^{^(-34)}} \\ \\ \\ f=1.95\text{ Hz.} \end{gathered}`

Therefore, the frequency of the pendulum is 1.95 Hz.

ANSWER:

1.95 Hz