Question:
On p. 15 of Bohr’s paper there is a discussion of the quantization of the angular momentum, $M$, of the electron orbiting an atomic nucleus. Bohr states that $M$ must be an integral multiple of a fundamental quantity, $M_0$. Bohr states a numerical value of $M_0$ in an equation on that page. How
much does his number differ from the presently accepted value of that quantity?
Note that Bohr does not explicitly state on that page what system of units he is using!
Solution:
It is quite obvious that on page 15 Bohr mentioned the quantity is $ 1.04 \times 10^{-27} $. The challenge, really, is to determine the unit of that value Bohr meant.
Today's value should be $ 1.05 \times 10^{-34} $. There is a 10 million fold changes, so the changes should be accounted for by the unit of measurement.
After messing up around online searching, I found the Gaussian units (or cgs units). If we apply the units to the planck's constant, we will find agreement.
Energy = Force $ \times $ distance = Mass $ \times $ acceleration $ \times $ distance = $ (kgms^{-2})m = kgm^2s^{-2} $.
Therefore, the Planck constant has unit energy $ \times $ time = $ (kgm^2s^{-2})s = kgm^2s^{-1} $. Now we substitute in the units using gram and cm instead, we will get a 1000 fold from the kg to gram conversion, and 10,000 fold from the $ m^2 $ to $ cm^2 $.
Wow - that explain the 10,000,000 fold of the value difference, now we see the values is surprisingly close, the relative error is less than 1%!
On p. 15 of Bohr’s paper there is a discussion of the quantization of the angular momentum, $M$, of the electron orbiting an atomic nucleus. Bohr states that $M$ must be an integral multiple of a fundamental quantity, $M_0$. Bohr states a numerical value of $M_0$ in an equation on that page. How
much does his number differ from the presently accepted value of that quantity?
Note that Bohr does not explicitly state on that page what system of units he is using!
Solution:
It is quite obvious that on page 15 Bohr mentioned the quantity is $ 1.04 \times 10^{-27} $. The challenge, really, is to determine the unit of that value Bohr meant.
Today's value should be $ 1.05 \times 10^{-34} $. There is a 10 million fold changes, so the changes should be accounted for by the unit of measurement.
After messing up around online searching, I found the Gaussian units (or cgs units). If we apply the units to the planck's constant, we will find agreement.
Energy = Force $ \times $ distance = Mass $ \times $ acceleration $ \times $ distance = $ (kgms^{-2})m = kgm^2s^{-2} $.
Therefore, the Planck constant has unit energy $ \times $ time = $ (kgm^2s^{-2})s = kgm^2s^{-1} $. Now we substitute in the units using gram and cm instead, we will get a 1000 fold from the kg to gram conversion, and 10,000 fold from the $ m^2 $ to $ cm^2 $.
Wow - that explain the 10,000,000 fold of the value difference, now we see the values is surprisingly close, the relative error is less than 1%!
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