A) 2 times the initial kinetic energy
B) 3 times the initial kinetic energy
C) 0.5 times the initial kinetic energy
D) 4 times the initial kinetic energy
Correct Answer: B
Solution :
: de Broglie wavelength of an electron having kinetic energy K is- \[\lambda =\frac{h}{\sqrt{2{{m}_{e}}K}}\]where \[{{m}_{e}}\] is the mass of an electron \[{{\lambda }^{2}}=\frac{{{h}^{2}}}{2{{m}_{e}}K}\] or \[K=\frac{h}{2{{m}_{e}}{{\lambda }^{2}}},\]\[K\propto \frac{1}{{{\lambda }^{2}}}\] \[\therefore \] \[\frac{K}{K}={{\left( \frac{\lambda }{\lambda } \right)}^{2}}={{\left( \frac{1nm}{0.5nm} \right)}^{2}}=\frac{1}{4}\] \[K=4K\] Additional energy given \[=K-K=4K-K=3K\]
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