Question

Chapter 29, Problem 53 An electron and a proton have the same kinetic energy and are moving at speeds much less than the speed of light. Determine the ratio of the de Broglie wavelength of the electron to that of the proton Nurnber Units

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Answer #1

Since the kinetic energies are the same

mev/2

Therefore

Ve/v mp/me ...(1)

The ratio of de Broglie wavelength is

lambda_e/lambda_p=m_p v_p/m_e v_e

Using the relation (1) one obtains

λ./λ,-V/rn p/ml.- V/ 1.673 × 10-27/9.11 × 10-31-42.815

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