Question

The de Broglie relation is obtained by equating

• A. $E={}^1/{}_{2}m{c}^2$ 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.