Question:medium

In a \(^1\mathrm{H}\) NMR spectrum obtained from a spectrometer operating at a magnetic field of \(14.1\ \mathrm{T}\), two resonances are observed at \(1.25\) ppm and \(5.75\) ppm. The separation between the two resonances (in Hz) is (rounded off to one decimal place).

(Given: gyromagnetic ratio (\(\gamma\)) of \(\mathrm{H} = 2.675\times10^{8}\ \mathrm{T^{-1}\,s^{-1}}\))

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Find the spectrometer operating (Larmor) frequency from \(\nu_0=\gamma B_0/2\pi\), then convert the ppm gap to Hz using \(\Delta\nu = \delta_{ppm}\times\nu_0(\mathrm{MHz})\).
Updated On: Jul 20, 2026
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Correct Answer: 2701

Solution and Explanation

A different way to reach the same number is to work out the absolute frequency of each peak first, then subtract, instead of scaling the ppm difference in one shot.

First get the spectrometer frequency from the proton's gyromagnetic ratio:

\[ \nu_0 = \frac{\gamma B_0}{2\pi} = \frac{(2.675\times10^8)(14.1)}{2\pi} = 6.0028\times10^8\ \mathrm{Hz} = 600.28\ \mathrm{MHz} \]

By definition, a peak at $\delta$ ppm sits $\delta\times10^{-6}$ of the operating frequency away from the reference, so its absolute frequency is $\nu(\delta) = \nu_0(1+\delta\times10^{-6})$.

For the peak at 1.25 ppm:

\[ \nu_1 = 600.28\ \mathrm{MHz} + 750.35\ \mathrm{Hz} \]

For the peak at 5.75 ppm:

\[ \nu_2 = 600.28\ \mathrm{MHz} + 3451.6\ \mathrm{Hz} \]

The separation is just the difference of the two added-on Hz terms, since the common 600.28 MHz cancels:

\[ \Delta\nu = 3451.6 - 750.35 = 2701.3\ \mathrm{Hz} \]

This matches scaling the $(5.75-1.25)=4.50$ ppm gap by $\nu_0$ directly, confirming that shortcut. The two peaks are $2701.3\ \mathrm{Hz}$ apart, well inside the accepted band of $2699$ to $2703\ \mathrm{Hz}$.

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