Discussion Forum

The energy that should be added to an electron  to reduce its de- Broglie wavelength from 1 nm to 0.5 nm is 

Answer:

Explanation:

 We know that,

 We have,   $\lambda=\frac{h}{\sqrt{2mk}}$

 $\therefore$    Kinetic energy (KE)   $\propto \frac{1}{\lambda^{2}}$

 Hence,   $\frac{KE_{2}}{KE_{1}}= \left(\frac{\lambda_{1}}{\lambda_{2}}\right)^{2}$

 Here, $\lambda_{1}$=1 nm

   $\lambda_{2}$=0.5 nm

$\frac{KE_{2}}{KE_{1}}= \left(\frac{1}{0.5}\right)^{2}$

KE₂ = 4KE₁

 Hence, the kinetic energy is increasing three times

 $\triangle KE=3KE_{1}$