If \( \lambda \) and \( K \) are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be:
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : The potential (V) at any axial point, at 2 m distance(r) from the centre of the dipole of dipole moment vector
\(\vec{P}\) of magnitude, 4 × 10-6 C m, is ± 9 × 103 V.
(Take \(\frac{1}{4\pi\epsilon_0}=9\times10^9\) SI units)
Reason R : \(V=±\frac{2P}{4\pi \epsilon_0r^2}\), where r is the distance of any axial point, situated at 2 m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below :
The output (Y) of the given logic gate is similar to the output of an/a :
One of the equations that are commonly used to define the wave properties of matter is the de Broglie equation. Basically, it describes the wave nature of the electron.
Very low mass particles moving at a speed less than that of light behave like a particle and waves. De Broglie derived an expression relating to the mass of such smaller particles and their wavelength.
Plank’s quantum theory relates the energy of an electromagnetic wave to its wavelength or frequency.
E = hν …….(1)
E = mc2……..(2)
As the smaller particle exhibits dual nature, and energy being the same, de Broglie equated both these relations for the particle moving with velocity ‘v’ as,
This equation relating the momentum of a particle with its wavelength is de Broglie equation and the wavelength calculated using this relation is the de Broglie wavelength.