A) \[E={\scriptstyle{}^{1}/{}_{2}}m{{c}^{2}}\]and \[E=h\upsilon \]
B) \[E=m{{c}^{2}}\]and \[E=h\upsilon \]
C) \[E={{p}^{2}}/2m\]and \[E=h\upsilon \]
D) none of these.
Correct Answer: B
Solution :
: \[E=m{{c}^{2}}\]and \[E=h\upsilon \] \[\Rightarrow \]\[m{{c}^{2}}=h\upsilon \]. Also\[\upsilon =c/\lambda \]. \[\therefore \]\[m{{c}^{2}}=\frac{hc}{\lambda }\Rightarrow mc=\frac{h}{\lambda }\Rightarrow \lambda =\frac{h}{mc}\]. de-Broglie assumed that the above relation also holds good for material particles like electrons. Hence for an electron,\[\lambda =h/mv,\]where v is its velocity.
You need to login to perform this action.
You will be redirected in
3 sec
