An electron,a doubly ionized helium ion (He^{ | vedclass

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(D) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(KE)}}$.Since the kinetic energy $(KE)$ and

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$\lambda_{e} > \lambda_{P} > \lambda_{He^{++}}$

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(D) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(KE)}}$.Since the kinetic energy $(KE)$ and Planck's constant $(h)$ are the same for all particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.Comparing the masses of the particles: $m_{He^{++}} > m_{P} > m_{e}$.Since $\lambda$ is inversely proportional to the square root of mass,the particle with the smallest mass will have the largest wavelength.Therefore,the relation is: $\lambda_{e} > \lambda_{P} > \lambda_{He^{++}}$.

An electron,a doubly ionized helium ion $(He^{++})$ and a proton have the same kinetic energy. The relation between their respective de-Broglie wavelengths $\lambda_{e}, \lambda_{He^{++}}$ and $\lambda_{P}$ is:

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