Proof of Bohr's postulate regarding angular momentum

May 12th, 2019 by Aziz Lokhandwala

Assumption: Assume a particle oscillating with displacement 𝑥. Now, $$ 𝑥 = 𝐴 sin(2π𝑓𝑡)$$

On differentiating above equation we get 'v' for that particle, $${d𝑥\over d𝑡} = 𝑣 = 2π𝑓𝐴 cos(2π𝑓𝑡) $$ Now 𝐾𝐸max can be given when cosine value is 1; thus, $$𝐾𝐸max = 2(π^2)(𝑓^2)𝑚(𝐴^2)... (i)$$ Now let linear momentum be given as: $$𝑝 = 2π𝑚𝑓𝐴 cos(2π𝑓𝑡)$$ Let 2π𝑚𝑓𝐴 = 𝐵 Now, $$sin^2(2π𝑓𝑡) + cos^2(2π𝑓𝑡) = 1$$ Therefore, $${𝑥^2 \over 𝐴^2} + {𝑝^2\over 𝐵^2} = 1$$ Now the area of this ellipse is given as π𝐴𝐵, therefore, $$∫𝑝d𝑥 = π𝐴𝐵$$ Now, $$𝐸 = 𝑛ℎ𝑓$$ Above equation is written using planck's postulate, thus by using equation (i) we have, $$2(π^2)(𝑓^2)𝑚(𝐴^2) = 𝑛ℎ𝑓 ...(ii)$$ Using (ii) we have, $$∫𝑝d𝑥 = 𝑛ℎ$$ Now, because we want to find angular momentum instead of linear momentum, we have, $$dx = rdθ$$ $$r * p = L$$ Integrating from 0 to 2π we have, $$𝐿= {𝑛ℎ \over2π}$$ Hence Proved