A photon of wavelength λ has energy E₀, equal to the kinetic energy of a proton of mass $m_p.$ The ratio $\lambda_{p}$ / $\lambda_{ph}$ of de Broglie wavelength of the proton to wavelength of the photon is:
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Question

A photon of wavelength λ has energy E₀, equal to the kinetic energy of a proton of mass $m_p.$ The ratio $\lambda_{p}$ / $\lambda_{ph}$ of de Broglie wavelength of the proton to wavelength of the photon is:

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G1 Physics · Accepted Answer

For the proton, E₀=p²/(2 m p ) , m_p), m p ​ ) , so λ_p=h/√(2 m p E 0 ) . m_pE₀). m p ​ E 0 ​ ) . For the photon, λ_ph=hc/E₀. Thus λ_p/λ_ph=(1/c)√(E₀/(2 m p ) ) . m_p)). m p ​ )) .

A photon of wavelength λ has energy E₀, equal to the kinetic energy of a proton of mass m p . m_p. m p ​ . The ratio λ p \lambda_{p} λ p ​ / λ p h \lambda_{ph} λ p h ​ of de Broglie wavelength of the proton to wavelength of the photon is:

A photon of wavelength λ has energy E₀, equal to the kinetic energy of a proton of mass m p . m_p. m p . The ratio λ p \lambda_{p} λ p / λ p h \lambda_{ph} λ p h of de Broglie wavelength of the proton to wavelength of the photon is: