A) 1.239 m
B) 0.149 m
C) 3.182 m
D) 2.33 m
Correct Answer: B
Solution :
[b] If \[{{v}_{\max }}\] is the speed of the fastest electron emitted from the metal surface, then \[\frac{hc}{\lambda }={{W}_{0}}+\frac{1}{2}mv_{\max }^{2}\] \[\frac{\left( 6.63\times {{10}^{-34}} \right)\times \left( 3\times {{10}^{8}} \right)}{\left( 180\times {{10}^{-9}} \right)}\] The radius of the electron is given by \[r=\frac{mv}{qB}=\frac{\left( 9.1\times {{10}^{-31}} \right)\times \left( 1.31\times {{10}^{6}} \right)}{\left( 1.6\times {{10}^{-19}} \right)\times \left( 5\times {{10}^{-9}} \right)}=0.149m\]
You need to login to perform this action.
You will be redirected in
3 sec
