1. **State the problem:** We have a radio receiver operating at 30°C detecting frequencies from 70 kHz to 130 kHz. We need to find: (i) The signal bandwidth. (ii) The thermal noise power density in Watts and in decibels (dB). 2. **Signal bandwidth formula:** The bandwidth $B$ is the difference between the maximum and minimum frequencies: $$B = f_{max} - f_{min}$$ 3. **Calculate bandwidth:** Given $f_{max} = 130$ kHz and $f_{min} = 70$ kHz, $$B = 130,000 - 70,000 = 60,000 \text{ Hz}$$ 4. **Thermal noise power density formula:** Thermal noise power density $N_0$ is given by: $$N_0 = kT$$ where $k$ is Boltzmann's constant and $T$ is the absolute temperature in Kelvin. 5. **Convert temperature to Kelvin:** $$T = 30 + 273.15 = 303.15 \text{ K}$$ 6. **Calculate thermal noise power density in Watts:** Given $k = 1.38 \times 10^{-23}$ J/K, $$N_0 = 1.38 \times 10^{-23} \times 303.15 = 4.18 \times 10^{-21} \text{ Watts/Hz}$$ 7. **Convert thermal noise power density to dB:** Using the formula: $$N_0(dB) = 10 \log_{10}(N_0)$$ $$N_0(dB) = 10 \log_{10}(4.18 \times 10^{-21}) = 10 \times (-20.38) = -203.8 \text{ dB}$$ **Final answers:** (i) Signal bandwidth $B = 60,000$ Hz (ii) Thermal noise power density $N_0 = 4.18 \times 10^{-21}$ Watts/Hz and $N_0(dB) = -203.8$ dB