A) \[102pm\]
B) \[124pm\]
C) \[95pm\]
D) \[112pm\]
Correct Answer: D
Solution :
: The de-Broglie wavelength of an electron is \[\lambda =\frac{h}{\sqrt{2mK}}\] where h is the Plancks constant, m is the mass of the electron and K is its kinetic energy, Here, \[h=6.63\times {{10}^{-34}}Js\] \[m=9.1\times {{10}^{-31}}kg\] \[K=120eV=120\times 16\times {{10}^{-19}}J\] \[(\because \,1eV=1.6\times {{10}^{-19}}J)\] \[\therefore \] \[\lambda =\frac{6.63\times {{10}^{-34}}Js}{\sqrt{2(9.1\times {{10}^{-31}}kg)(120\times 1.6\times {{10}^{-19}}J)}}\] \[=1.12\times {{10}^{-10}}m=112\times {{10}^{-12}}m\] \[=112pm\] (\[\because \] \[1pm={{10}^{-12}}m\])
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