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  • The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is ve = c/100. Then (a) Ee/Ep = 10^{-4}(b) Ee/Ep = 10^{-2}

The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is ve = c/100. Then

(a) Ee/Ep = 10-4

(b) Ee/Ep = 10-2

(c) Pe/mec = 10-2

(d) Pe/mec = 10-4

Answers (1)

The answer is the option (b,c)
The energy of a charged particle when accelerated through a potential difference V is:

E=\frac{1}{2}mv^{2}=qV

De-Broglie wavelength

\lambda=\frac{h}{p}=\frac{h}{\sqrt{2mE}}=\frac{h}{\sqrt{2mqV}}\\ \lambda_{Neutron}=\frac{0.286*10^{-10}}{\sqrt{E}}m=\frac{0.286*10^{-10}}{\sqrt{E}}A^{\circ}

At ordinary temperature, the energy of thermal neutrons

E=kT\Rightarrow \lambda=\frac{h}{\sqrt{2mkT}}\\ \lambda_{Thermal\; Neutron}=\frac{30.83}{\sqrt{T}}A^{\circ}\\ Mass\; of\; electron=m_{e}\\ Mass\; of\; photon=m_{p}\\

de Broglie wavelength of the electron

\lambda_{e}=\frac{h}{m_{e}v_{e}}=\frac{100h}{m_{e}c}\\ KE_{e}=\frac{1}{2}m_{e}v_{e}^{2}\\ m_{e}v_{e}^{2}=\sqrt{2E_{e}m_{e}}\\ \lambda_{e}=\frac{h}{m_{e}v_{e}}=\frac{h}{\sqrt{2E_{e}m_{e}}}\\ E_{e}=\frac{h^{2}}{2\lambda_{e}^{2}v_{e}}

de Broglie wavelength of the proton is \lambda_{p}
E_{p}=\frac{hc}{\lambda_{p}}=\frac{hc}{2\lambda_{e}}\\ \frac{E_{p}}{E_{e}}=\left ( \frac{hc}{\lambda_{e}} \right )*\left ( \frac{2\lambda_{e}^{2}m_{e}}{h} \right )=100\\ p_{e}=m_{e}v_{e}=m_{e}*\frac{c}{100}\\ \frac{p_{e}}{m_{e}c}=10^{-2}

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