A) 1
B) 1836
C) 1836
D) 918
Correct Answer: B
Solution :
de-Broglie's wavelength, \[\lambda =\frac{h}{mv}\] For electron, \[{{\lambda }_{e}}=\frac{h}{{{m}_{e}}v}\] For photon, \[{{\lambda }_{P}}=\frac{h}{{{m}_{p}}v}\] \[\therefore \] \[\frac{{{\lambda }_{e}}}{{{\lambda }_{p}}}=\frac{h/{{m}_{e}}v}{h/{{m}_{p}}v}\] \[\frac{{{\lambda }_{e}}}{{{\lambda }_{p}}}=\frac{{{m}_{p}}}{{{m}_{e}}}\] \[\frac{{{\lambda }_{e}}}{{{\lambda }_{p}}}=\frac{1.67\times {{10}^{-27}}}{9.1\times {{10}^{-31}}}=0.18\times {{10}^{4}}=1836\]
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