A) \[+3.4eV,0.66\times {{10}^{-9}}m\]
B)
C) \[2.8eV,2.38\times {{10}^{-10}}m\]
D) \[1.1eV,1.28\times {{10}^{-9}}m\]
Correct Answer: A
Solution :
[a] \[{{E}_{n}}=-3.4eV\] The kinet2ic energy is equal to the magnitude of total energy in this case. \[\therefore K.E.=+3.4eV\] The de Broglie wavelength of electron \[\lambda =\frac{h}{\sqrt{2mK}}=\frac{6.64\times {{10}^{-34}}}{\sqrt{2\times 9.1\times {{10}^{-31}}\times 3.4\times 1.6\times {{10}^{-19}}}}\]\[=0.66\times {{10}^{-9}}m\]
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