The de Broglie wavelengths associated with an electron and a proton is

1.	The de Broglie wavelengths associated with an electron and a proton is equal. Prove that the kinetic energy of the electron is greater than that of the proton.
Answer» de Brogie wavelength λ = \(\frac{h}{\sqrt {2mK}}\) We get K = \(\frac{h^2}{{2mλ^2}}\) K ∝ \(\frac{1}{m}\) We know that m_p > m_e _{\(\frac{K_e}{K_p} = \frac {m_p}{m_e} > 1\)} K_e > K_p

The de Broglie wavelengths associated with an electron and a proton is equal. Prove that the kinetic energy of the electron is greater than that of the proton.

Answer»

de Brogie wavelength λ = \(\frac{h}{\sqrt {2mK}}\)

We get K = \(\frac{h^2}{{2mλ^2}}\)

K ∝ \(\frac{1}{m}\)

We know that m_p > m_e

_{\(\frac{K_e}{K_p} = \frac {m_p}{m_e} > 1\)}

K_e > K_p

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The de Broglie wavelengths associated with an electron and a proton is equal. Prove that the kinetic energy of the electron is greater than that of the proton.

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