Question

The de Broglie relation is obtained by equating

• 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.

Answer 1

Answer: (B)

: $ 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.