The de-Broglie wavelength of a particle having kinetic energy E is λ . How much extra energy must be…

Recall the formula of de-Broglie wavelength, $\lambda =\frac{h}{p}$, here $p$ is the linear momentum of the particle, now use the relation between linear momentum and kinetic energy, $p=\sqrt{2Em}\Rightarrow E\propto \frac{1}{{\lambda }^2}$$\Rightarrow E=\frac{k}{{\lambda }^2}$

$$\begin{gathered} \text{and} E^{\text{'}}=\frac{16k}{9{\lambda }^2} \\ \Rightarrow \Delta E=\frac{7E}{9} \end{gathered}$$.