Question

What will be the kinetic energy of an electron having de-Broglie wavelength $2\stackrel{\circ }{A}$?

• A. $37.5eV$
• B. $75eV$
• C. $150eV$
• D. $300eV$

Answer 1

The correct option is A $37.5eV$

The de-Broglie wavelength associated with electron
${\lambda }_e=\frac{1.227}{\sqrt{V}}nm$
We get $V=\frac{(12.27{)}^2}{{\lambda }_e^2}\times {10}^{-2}$
$\Rightarrow V=\frac{150}{{\lambda }_e^2}\times {10}^{-2}$

where ${\lambda }_e=0.2nm$
$\Rightarrow V=\frac{150\times {10}^{-2}}{(0.2{)}^2}$
$\Rightarrow V=\frac{150\times {10}^{-2}}{4\times {10}^{-2}}=37.5$ volts
$\therefore$ Kinetic energy $E=37.5eV$.