Question

If the frequency of the 13C signal of TMS is 201.16 MHz, the two 13C signals of acetic acid at 179.0 and 20.0 ppm are separated by a frequency difference in kHz (in decimal form with two decimal places) equal to

Answer 1

The difference in frequency of the two signals is $1.33 \times 10^{10} \ kHz$.

The given parameters;

• frequency of the 13 C signal = 201.16 MHz

The energy of the 13 C signal located at 20 ppm is calculated as follows;

$E = hf\\\\E_1 = h \frac{c}{\lambda} \\\\E_1 = \frac{(6.626 \times 10^{-34})\times 3\times 10^8}{20 \times 10^{-6}} \\\\E_1 = 9.94 \times 10^{-21} \ J$

The energy of the 13 C signal located at 179 ppm is calculated as follows;

$E_2 = \frac{hc}{\lambda} \\\\E_2 = \frac{(6.626\times 10^{-34})\times (3\times 10^{8})}{179 \times 10^{-6} } \\\\E_2 = 1.11 \times 10^{-21} \ J$

The difference in frequency of the two signals is calculated as follows;

$E_1- E_2 = hf_1 - hf_2\\\\E_1 - E_2 = h(f_1 - f_2)\\\\f_1 - f_2 = \frac{E_1 - E_2 }{h} \\\\f_1 - f_2 = \frac{(9.94\times 10^{-21}) - (1.11 \times 10^{-21})}{6.626\times 10^{-34}} \\\\f_1 - f_2 = 1.33 \times 10^{13} \ Hz\\\\f_1 - f_2 = 1.33\times 10^{10} \ kHz$

Thus, the difference in frequency of the two signals is $1.33 \times 10^{10} \ kHz$.

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