The energy that should be added to an electron, to reduce its deBroglie wavelength from  $2\times {10}^{-9}\text{\hspace{0.17em}m}$ to $0.5\times {10}^{-9}\text{\hspace{0.17em}m}$ will be______(in eV) ($h=6.6\times {10}^{-34}\text{\hspace{0.17em}Js}$ ,  $m_e=9\times {10}^{-31}\text{\hspace{0.17em}kg}$)

Given data
$h=$ Plank’s constant $=6.6\times {10}^{-3}\text{\hspace{0.17em}Js}$
$m_e=$ Mass of electron= $9\times {10}^{-31}\text{\hspace{0.17em}kg}$
We use De-broglie wavelength $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}\left(\because E=\frac{p^2}{2m}\right)$ 
$\Rightarrow E=\frac{h^2}{2m{\lambda }^2}$ 
Thus we have   $\Delta E=\frac{h^2}{2m}\left(\frac{1}{{\lambda }_1^2}-\frac{1}{{\lambda }_2^2}\right)$
Using ${\lambda }_1=0.5\times {10}^{-9}\text{\hspace{0.17em}m}$ and  ${\lambda }_2=2\times {10}^{-9}\text{\hspace{0.17em}m}$

$\begin{array}{l}\mathrm{\Delta }E=\frac{{\left(6.6\times {10}^{-34}\right)}^2}{2\times 9\times {10}^{-31}}\left[\frac{1}{{\left(0.5\times {10}^{-9}\right)}^2}-\frac{1}{{\left(2\times {10}^{-9}\right)}^2}\right]\\ \mathrm{\Delta }E=2.42\times {10}^{-37}[4-0.25]\times {10}^{18}=9.075\times {10}^{-19}J\\ =\frac{9.075\times {10}^{-19}}{1.6\times {10}^{-19}}\mathrm{eV}\left(\because 1ev=1.6\times {10}^{-19}J\right)\\ \therefore \mathrm{\Delta }E=5.67\mathrm{eV}\end{array}$