A) 0.61 m
B) 0.61 cm
C) \[6.1\times {{10}^{-4}}m\]
D) \[6.1\times {{10}^{-5}}m\]
Correct Answer: A
Solution :
: For photoelectron emission, \[\frac{hc}{\lambda }=\frac{1}{2}m{{v}^{2}}\] For circular motion of electron in magnetic field B, \[Bev=\frac{m{{v}^{2}}}{r}\] or \[Bev=\frac{2hc}{\lambda r}\]or\[r=\frac{2hc}{\lambda Bev}\] \[r=\frac{2\times (6.6\times {{10}^{-34}})\times (3\times {{10}^{8}})}{(2537\times {{10}^{-10}})\times (1.6\times {{10}^{-19}})\times (1.6\times {{10}^{5}})}\] \[=\frac{2\times 6.6\times 3\times 100}{2537\times 1.6\times 1.6}\] \[\therefore \]radius = 0.61 m.
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