Use uncertainty principle to obtain the result of Bohr's Model

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SUMMARY

The forum discussion focuses on calculating the minimum energy of the hydrogen atom using the uncertainty principle. The user initially derives an incorrect energy value of -54.7 eV instead of the expected -13.6 eV by applying the uncertainty relation Δr Δp ≥ h/4π. The error arises from misapplying the uncertainty in position Δr and momentum Δp, particularly in the treatment of the atom as a 1-D system. The correct approach involves adjusting Δr to (1/2)Δr, which aligns with the established results of Bohr's Model.

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HAMJOOP
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Problem
Find the minimum energy of the hydrogen atom by using uncertainty principle

a. Take the uncertainty of the position Δr of the electron to be approximately equal to r
b. Approximate the momentum p of the electron as Δp
c. Treat the atom as a 1-D system


My step

1. Δr Δp ≥ h/4(pi)
Δp ≥ h/4(pi)r

2. Total energy E = (p^2 /2m) - ke^2 /r

≥ (h^2/8(pi)^2 m r^2) - (ke^2)/r

3. rearranging the term

(Emin)r^2 + (ke^2)r - (h^2 /8(pi)^2 m ) = 0

Require Δ = 0 for the quadratic equation

I obtain E = -54.7 eV ≠ -13.6 eV


If I replace Δr by (1/2)Δr, I can obtain the correct result. But I don't know why.
 
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HAMJOOP said:
1. Δr Δp ≥ h/4(pi)
Δp ≥ h/4(pi)r

2. Total energy E = (p^2 /2m) - ke^2 /r

≥ (h^2/8(pi)^2 m r^2) - (ke^2)/r
you have p=h/4(pi)r, but then you seem to say p^2 /2m = h^2/8(pi)^2 m r^2 there is a mistake in this step
 
woah there! you can't shift Δr to the RHS in case there is a > sign,right? if Δr.Δp=h/4π, ⇒Δp=h/4πr
 

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