Question

The de-Broglie wavelength of a particle having kinetic energy E is $\lambda$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to $75\%$ of the initial value ?

• A. $\frac{16}{9}E$
• B. $E$
• C. $\frac{7}{9}E$
• D. $\frac{1}{9}E$

Answer 1

The correct option is C $\frac{7}{9}E$

Since, de-Broglie wavelength in terms of kinetic energy E is given by,

$\lambda =\frac{h}{\sqrt{2mE}}$

It means wavelength of the particle is inversely proportional to square root of kinetic energy.

i.e. $\lambda \propto \frac{1}{\sqrt{E}}$

Since, ${\lambda }_2=0.75{\lambda }_1$

So, $\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{E_2}{E_1}}=\frac{4}{3}$,

Therefore, $\frac{E_2}{E_1}=\frac{16}{9}$

So the extra energy which is required in this case is,

$E=\frac{16}{9}{E}_1-E_1=\frac{7}{9}{E}_1=\frac{7}{9}E$