The de-Broglie wavelength of a particle having kinetic energy

$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}, \mathrm{mv}=\sqrt{2 \mathrm{mE}}$

$\lambda \propto \frac{1}{\sqrt{E}}$

$\frac{\lambda_{2}}{\lambda_{1}}=\sqrt{\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}}=\frac{3}{4}, \lambda_{2}=0.75 \lambda_{1}$

$\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{3}{4}\right)^{2}$

$\mathrm{E}_{2}=\frac{16}{9} \mathrm{E}_{1}=\frac{16}{9} \mathrm{E}$

$\left(\mathrm{E}_{1}=\mathrm{E}\right)$

Extra energy given $=\frac{16}{9} \mathrm{E}-\mathrm{E}=\frac{7}{9} \mathrm{E}$

Ans. 2