Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 36A / b / i

Solution

h^{2}

Z_{1} = \frac{1}{h ^{2}} \int_{R^{2}} e^{- β p^{2} / (2 m)} d^{2} p \int_{r < R} e^{- β κ r^{2} /2} d^{2} x = \frac{1}{h ^{2}} (\frac{2 πm}{β}) (2 π \int_{0}^{R} r e^{- β κ r^{2} /2} d r) = \frac{4 π ^{2} m}{h ^{2} β ^{2} κ} (1 - e^{- β κ R^{2} /2}) .

Equivalently, using

h = 2 π ℏ

Z_{1} = \frac{m}{ℏ ^{2} β ^{2} κ} (1 - e^{- β κ R^{2} /2}) .

There are no discussions about this article yet.

There are currently no matching articles.