A) four times the initial energy
B) thrice the initial energy
C) equal to the initial energy
D) twice the initial energy
Correct Answer: A
Solution :
Initial de-Broglie wavelengths \[{{\lambda }_{1}}={{10}^{-10}}m\] Final de-Broglie wavelength \[{{\lambda }_{2}}=0.5\times {{10}^{-10}}m\] The velocity of an electron is given by \[\upsilon =\frac{h}{m\lambda }\] The kinetic energy of an electron \[E=\frac{1}{2}m{{\upsilon }^{2}}=\frac{1}{2}m{{\left( \frac{h}{m\lambda } \right)}^{2}}\] \[=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}\propto \frac{1}{{{\lambda }^{2}}}\] Hence,\[\frac{{{E}_{2}}}{{{E}_{1}}}={{\left( \frac{{{\lambda }_{1}}}{{{\lambda }_{2}}} \right)}^{2}}={{\left( \frac{{{10}^{-10}}}{0.5\times {{10}^{-10}}} \right)}^{2}}=4\]
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