Calculate the amount of energy (in calories) required to flip an ¹Hnucleus in an NMR spectrometer that operates at 300 MHz.

NMR spectroscopy identifies the carbon-hydrogen framework of an organic compound. The greater the magnetic field produced by the magnet, the greater the operating frequency of the spectrophotometer.

\[E = h\nu \]

Where E is the energy required, h is Planck’s constant, \(\left( {6.627 \times {{10}^{ - 34}}J - S} \right)\) and \(\nu \) is the frequency 300 MHz \(\left( {300 \times {{10}^6}\,{s^{ - 1}}} \right)\).

Upon substituting all the known values in the given equation, we get,

\(\begin{array}{c}E = h\nu \\ = 6.627 \times {10^{ - 34}}J - S \times 300 \times {10^6}\,{s^{ - 1}}\\ = 1988.1 \times {10^{ - 28}}J\end{array}\)

The standard relationship between joule and calorie is as follows:

\(1\,J = 0.2390057\,calories\)

Therefore, \(1988.1 \times {10^{ - 28}}J\) is converted to calories as follows:

\(\begin{array}{c}1\,J = 0.2390057\,calories\\1988.1 \times {10^{ - 28}}J = 1988.1 \times {10^{ - 28}} \times 0.2390057\,calories\\ = 475.16 \times {10^{ - 28}}\,calories\end{array}\)

Thus, the amount of energy required to flip is \(475.16 \times {10^{ - 28}}\,calories\) .