A sub-atomic particle of mass \( 10^{-30} \) kg is moving with a velocity of \( 2.21 \times 10^6 \) m/s. Under the matter wave consideration, the particle will behave closely like (h = \( 6.63 \times 10^{-34} \) J.s)
To determine the type of radiation with which the sub-atomic particle behaves closely, we need to calculate its de Broglie wavelength. This is given by the equation:
\(\lambda = \frac{h}{mv}\)
where:
Substituting the values into the de Broglie wavelength equation:
\(\lambda = \frac{6.63 \times 10^{-34}}{10^{-30} \times 2.21 \times 10^6}\)
Solving for \( \lambda \):
\(\lambda = \frac{6.63 \times 10^{-34}}{2.21 \times 10^{-24}}\)
\(\lambda = 3 \times 10^{-10} \) m
This wavelength corresponds to the X-ray region of the electromagnetic spectrum, which typically ranges from \( 10^{-11} \) m to \( 10^{-8} \) m. As a result, the sub-atomic particle behaves like X-rays.
Thus, the correct answer is: X-rays.
Match List-I with List-II on the basis of two simple harmonic signals of the same frequency and various phase differences interacting with each other:
LIST-I (Lissajous Figure) | LIST-II (Phase Difference) | ||
---|---|---|---|
A. | Right handed elliptically polarized vibrations | I. | Phase difference = \( \frac{\pi}{4} \) |
B. | Left handed elliptically polarized vibrations | II. | Phase difference = \( \frac{3\pi}{4} \) |
C. | Circularly polarized vibrations | III. | No phase difference |
D. | Linearly polarized vibrations | IV. | Phase difference = \( \frac{\pi}{2} \) |
Choose the correct answer from the options given below:
The motion of an airplane is represented by the velocity-time graph as shown below. The distance covered by the airplane in the first 30.5 seconds is km.
The least acidic compound, among the following is
Choose the correct set of reagents for the following conversion: