Step 1: Understanding the Concept:
The electromagnetic (EM) spectrum is the range of all types of EM radiation. Radiation is classified by wavelength or frequency. The order of the spectrum in terms of increasing frequency (and decreasing wavelength) is: Radio waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, Gamma rays.
Step 2: Detailed Explanation:
Let's analyze the given frequency ranges:
\begin{itemize}
\item Visible Light: The visible spectrum, which our eyes can see, ranges from approximately \(4.3 \times 10^{14}\) Hz (red) to \(7.5 \times 10^{14}\) Hz (violet).
\item Ultraviolet (UV) Radiation: UV radiation has a frequency just higher than that of violet light. Its range typically starts from the end of the visible spectrum and extends upwards. The accepted range for UV is roughly \(8 \times 10^{14}\) Hz to \(3 \times 10^{16}\) Hz.
\end{itemize}
(A) \(3 \times 10^{10} - 3 \times 10^{12}\) Hz: This corresponds to the microwave and far-infrared region.
(B) \(3 \times 10^{12} - 3 \times 10^{14}\) Hz: This corresponds to the infrared region.
(C) \(3 \times 10^{14} - 3 \times 10^{16}\) Hz: This range begins at the boundary of infrared/visible light and extends through the entire ultraviolet region, up to the start of the X-ray region. This is the most appropriate range among the choices.
(D) \(3 \times 10^{16}\) and above: Frequencies above \(3 \times 10^{16}\) Hz are generally classified as X-rays and Gamma rays.
Step 3: Final Answer:
The frequency range that best represents ultraviolet radiation from the given options is \(3 \times 10^{14} - 3 \times 10^{16}\) Hz. Therefore, option (C) is correct.
The path of scattered \( \alpha \)-particle is:
The maximum focal length of convex lens is for:
The power consumed in alternating current in a circuit containing only a capacitor will be:
The amplitude of the magnetic field of an electromagnetic wave in vacuum is \(B_0 = 510 \, \text{nT}\). What is the amplitude of the electric field of the wave?
What do you mean by the current sensitivity of a moving coil galvanometer? Resistance of a galvanometer is \( 50 \, \Omega \) and for full-scale deflection, the current is \( 0.05 \, \mathrm{A} \). What would be the required length of a wire to convert it into an ammeter of 5 A range? (Area of cross-section of wire = \( 2.7 \times 10^{-6} \, \mathrm{m^2} \), specific resistance of the wire material = \( 5.0 \times 10^{-7} \, \Omega \cdot \mathrm{m} \))