Question

The additional energy that should be given to an electron to reduce its de-Broglie wavelength from $1$nm to $0.5$nm is:

• A. $2$ times the initial kinetic energy
• B. $3$ times the initial kinetic energy
• C. $0.5$ times the initial kinetic energy
• D. $4$ times the initial kinetic energy

Answer 1

The correct option is B $3$ times the initial kinetic energy

From $\lambda =\frac{h}{p}$ where $p=\sqrt{2mK}$

We get kinetic energy $K=\frac{h^2}{2m{\lambda }^2}$

Initial kinetic energy $K_i=\frac{h^2}{2m}$ $(\because {\lambda }_i=1nm)$

Change in kinetic energy $\Delta K=\frac{h^2}{2m}(\frac{1}{{\lambda }_f^2}-\frac{1}{{\lambda }_i^2})$

$\therefore$ $\Delta K=\frac{h^2}{2m}(\frac{1}{(0.5{)}^2}-\frac{1}{1^2})=3\frac{h^2}{2m}=3K_i$

Hence additional kinetic energy to be added is 3 times the initial kinetic energy.