Answer:
(i) As both, electron and photon have same wavelength, so they have same momentum also. \[\therefore \] \[{{p}_{e}}:{{p}_{p}}=\mathbf{1:1}\] (ii) From the solution of Problem 30, \[\frac{\text{Energy of photon}}{\text{K}\text{.E of electron}}=\frac{2m\lambda c}{h}\] \[=\frac{2\times 9.11\times {{10}^{-31}}\times {{10}^{-9}}\times 3\times {{10}^{8}}}{6.6\times {{10}^{-34}}}\] \[=\frac{9110}{11}=\mathbf{9110:11}\]
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