Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
The difference in frequency of the two signals is [tex]1.33 \times 10^{10} \ kHz[/tex].
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;
[tex]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[/tex]
The energy of the 13 C signal located at 179 ppm is calculated as follows;
[tex]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[/tex]
The difference in frequency of the two signals is calculated as follows;
[tex]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[/tex]
Thus, the difference in frequency of the two signals is [tex]1.33 \times 10^{10} \ kHz[/tex].
Learn more here:https://brainly.com/question/14016376
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.