A) \[2.73\times {{10}^{-24}}\]
B) \[2.42\times {{10}^{-10}}\]
C) \[242.2\times {{10}^{10}}\]
D) None of these
Correct Answer: B
Solution :
One percent of the speed of light is \[v=\left( \frac{1}{100} \right)\,\,(3.00\times {{10}^{8}}m{{s}^{-1}})\] = \[3.00\times {{10}^{6}}m{{s}^{-1}}\] Momentum of the electron \[(p)\] = \[m\nu \] = \[(9.11\times {{10}^{-31}}kg)\,(3.00\times {{10}^{6}}m{{s}^{-1}})\] = \[2.73\times {{10}^{-24}}kg\,m{{s}^{-1}}\] The de-broglie wavelength of this electron is \[\lambda =\frac{h}{p}=\frac{6.626\times {{10}^{-34}}}{2.73\times {{10}^{-24}}kgm{{s}^{-1}}}\] \[\lambda =2.424\times {{10}^{-10}}m\].
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