Question

How will $\mathit{\Delta }\mathit{E}$for an isolated $13C$nucleus compare with that of a $1\mathrm{H}$nucleus?

Short answer

The required $\mathit{\Delta }\mathit{E}$for an isolated $13c$is $0.251506429$times higher than the $1H_{}$nucleus.

Step-by-step solution

Concept introduction

Nuclear magnetic resonance is a technique for predicting the compound's structural formula. The nucleus is examined using NMR spectroscopy in the presence of an external magnetic field.

Apply the formula

Let us use the formula :

$\frac{N_T}{N_0}=\exp \left(\frac{-\Delta E}{KT}\right)-----\left(\mathrm{I}\right)$

and also

$\frac{N_T}{N_0}=\exp \left(\frac{-\gamma h{B}_0}{2\pi KT}\right)$

$\frac{N_T}{N_0}=\exp \left(\frac{\frac{\gamma h{B}_0}{2\pi }}{KT}\right)-----\left(\mathrm{II}\right)$

Comparing the formulas

When I and II are compared, we get

$\Delta E=\frac{-\gamma h{B}_0}{2\pi }$

Hence,$\frac{\left(\Delta E\right)}{}$_C13(ΔE)1H=6.7283×107T-1s-1hB02π2.6752×108T-1s-1hB02π

$\begin{array}{l}=\frac{6.7283\times {10}^7}{2.6752\times {10}^8}\\ =0.251506429\end{array}$