A) 1 : 50
B) 1 : 20
C) 20 :1
D) 50 : 1
Correct Answer: C
Solution :
de-Broglie wavelength \[\lambda =\frac{h}{\sqrt{2mK}}\] The kinetic energy of the electron \[{{K}_{electron}}=\frac{1}{2m}.\frac{{{h}^{2}}}{{{\lambda }^{2}}}\] ?(i) where \[h=\]Planck constant \[\lambda =\]wavelength The photon energy \[{{E}_{photon}}=\frac{hc}{\lambda }\] ?(ii) From Eqs. (ii) and (i), we get \[\frac{{{E}_{photon}}}{{{K}_{electron}}}=\frac{hc/\lambda }{{{h}^{2}}/2m.{{\lambda }^{2}}}=\frac{hc.{{\lambda }^{2}}\times 2m}{{{h}^{2}}.\lambda }\] Where \[m=0.5\,MeV=5\times {{10}^{5}}eV\] \[\frac{h}{\lambda c}=50\times {{10}^{3}}eV=\frac{2m\lambda C}{h}\] \[=\frac{2m}{h/\lambda c}=\frac{2\times 5\times {{10}^{5}}}{50\times {{10}^{3}}}\] \[\left( \because \,m=\frac{h}{c\lambda } \right)\] \[=20:1\]
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