Question

A proton and an electron are accelerated by same potential difference starting from rest have de- Brogile wavelength ${\lambda }_p$ and ${\lambda }_{e^{\ast }}$

• A. ${\lambda }_e={\lambda }_p$
• B. ${\lambda }_0<{\lambda }_p$
• C. ${\lambda }_e>{\lambda }_p$
• D. none of these

Answer 1

Answer: (C)

${\lambda }_e>{\lambda }_p$

The wavelength of the photon is given as:

$\begin{array}{c}\lambda =\frac{h}{P}\\ =\frac{h}{\sqrt{2mk}}\\ \lambda \propto \frac{1}{\sqrt{m}}\\ \therefore {\lambda }_p<{\lambda }_e\\ Hence,optionCiscorrectanswer.\end{array}$