A) \[{{\left[ \frac{2(h\lambda -\theta )}{m} \right]}^{{1}/{2}\;}}\]
B) \[{{\left[ \frac{2(hc-\lambda \phi )}{m\lambda } \right]}^{{1}/{2}\;}}\]
C) \[\left[ \frac{2(hc-\lambda \phi )}{m} \right]\]
D) \[{{\left[ \frac{2(hc-\lambda \phi )}{m\lambda } \right]}^{{1}/{2}\;}}\]
Correct Answer: B
Solution :
From Einsteins photo electric equation. Energy of photon \[=\phi +\frac{1}{2}m{{\upsilon }^{2}}\] \[\frac{hc}{\lambda }=\phi +\frac{1}{2}m{{\upsilon }^{2}}\] \[\frac{1}{2}m{{\upsilon }^{2}}=\left( \frac{hc}{\lambda }=\phi \right)\] \[{{\upsilon }^{2}}=2\left( \frac{hc-\lambda \phi }{m\lambda } \right)\] \[\upsilon ={{\left[ \frac{2(hc-\lambda \phi )}{m\lambda } \right]}^{1/2}}\]
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