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Chapter 35: Problem 8

What did Einstein mean by his remark, loosely paraphrased, that "God does not play dice"?

Short Answer

Expert verified
Einstein's 'God does not play dice' was a critique of quantum physics, arguing that the universe is based on certain unchanging laws, not randomness or chance. He believed that the Universe operates under deterministic laws that aren't subjected to randomness.

Step by step solution

01

Explore the Context

Albert Einstein, one of the greatest physicists of all time, initially stated, 'God does not play dice' in response to the emerging field of quantum mechanics, which fundamentally shook established laws of physics. Quantum mechanics introduced uncertainty and randomness in the behavior of particles, a concept Einstein found difficult to accept.
02

Explain the Mean

Einstein's feelings about God not playing dice referred to his belief that the Universe, as God's creation, operates based on certain laws and rules that don't change and aren't left to chance. Hence, when quantum mechanics proposed that certain events occur randomly without specific determinable causes, Einstein challenged it.
03

Interpret the Quote

In a nutshell, with 'God does not play dice,' Einstein is voicing his position that the Universe follows deterministic laws, and events aren't simply random or by chance. He was expressing his rejection of the philosophical implications of quantum mechanics, which subvert classical, deterministic physics.

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Most popular questions from this chapter

The wave functions of Problem \(58,\) as well as their derivatives, need to be continuous at \(x=L\) if these functions are to represent the quantum state of a particle in the finite square well. (a) Show that these conditions lead to two equations: $$ \begin{array}{c} A \sin (\sqrt{\epsilon})=B e^{-\sqrt{\mu-\epsilon}} \\ \sqrt{\epsilon} A \cos (\sqrt{\epsilon})=-\sqrt{\mu-\epsilon} B e^{-\sqrt{\mu-\epsilon}} \end{array} $$ (b) then show that these lead to the single equation $$ \tan (\sqrt{\epsilon})=-\sqrt{\frac{\epsilon}{\mu-\epsilon}} $$

The table below lists the wavelengths emitted as electrons in identical square-well potentials drop from various states \(n\) to the ground state. Determine a quantity that, when you plot \(\lambda\) against it, should yield a straight line. Make your plot, establish a best-fit line, and use your line to determine the width of the square well. $$\begin{array}{|l|c|c|c|c|c|} \hline \text { Initial state, } n & 4 & 5 & 7 & 8 & 10 \\ \hline \text { Wavelength, } \lambda(\mathrm{nm}) & 1110 & 674 & 354 & 281 & 169 \\ \hline \end{array}$$

An alpha particle (mass 4 u) is trapped in a uranium nucleus with diameter 15 fm. Treating the system as a one-dimensional square well, what would be the minimum energy for the alpha particle?

A particle is confined to a 1.0 -nm-wide infinite square well. If the energy difference between the ground state and the first excited state is \(1.13 \mathrm{eV},\) is the particle an electron or a proton?

An electron is in a narrow molecule \(4.4 \mathrm{nm}\) long, a situation that approximates a one-dimensional infinite square well. If the electron is in its ground state, what is the maximum wavelength of electromagnetic radiation that can cause a transition to an excited state?
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