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:
The de-Broglie wavelength of an electron is the same as that of a photon. If the velocity of the electron is 25% of the velocity of light, then the ratio of the K.E. of the electron to the K.E. of the photon will be:
The de Broglie wavelengths of a proton and an α particle are \( \lambda \) and \( 2\lambda \) respectively. The ratio of the velocities of proton and α particle will be:
Electron beam used in an electron microscope, when accelerated by a voltage of $20 kV$, has a de-Broglie wavelength of $\lambda_0$. If the voltage is increased to $40 kV$, then the de-Broglie wavelength associated with the electron beam would be:
An electron (mass m) with an initial velocity\(\vec{v}=v_0\hat{i}(v_0>0)\)is moving in an electric field\(\vec{E}=E_0\hat{i}(E_0>0)\)where E0 is constant. If at t = 0 de Broglie wavelength is\(λ_0=\frac{ℎ}{mv_0}\), then its de Broglie wavelength after time t is given by