The kinetic energy of an electron, an α-particle, and a proton are given as 4K, 2K, and K respectively. The de…

Question

The kinetic energy of an electron, an $$\alpha$$-particle, and a proton are given as **$$4K$$**, **$$2K$$**, and **$$K$$** respectively. The de-Broglie wavelength associated with electron ($$\lambda_e$$), $$\alpha$$-particle ($$\lambda_{\alpha}$$), and the proton ($$\lambda_p$$) are as follows:

Accepted Answer

D. $$\lambda_{\alpha} < \lambda_p < \lambda_e$$

Answer

A. $$\lambda_{\alpha} > \lambda_p > \lambda_e$$

Answer

B. $$\lambda_{\alpha} = \lambda_p > \lambda_e$$

Answer

C. $$\lambda_{\alpha} = \lambda_p < \lambda_e$$

Answer & step-by-step solution

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