1. For a non-relativistic particle, de Broglie wavelength is

$\lambda = \frac{h}{p}$

and kinetic energy is

$E = \frac{p^2}{2m}$

So momentum is

$p = \sqrt{2mE}$

Hence,

$\lambda = \frac{h}{\sqrt{2mE}} \propto \frac{1}{\sqrt{E}}$

2. Let the new kinetic energy be

$E' = E + \Delta E$

The new wavelength is given to be

$\lambda' = \frac{\lambda}{2}$

Using $\lambda \propto 1/\sqrt{E}$,

$\frac{\lambda'}{\lambda} = \sqrt{\frac{E}{E'}}$

Substitute $\lambda' = \lambda/2$:

$\frac{1}{2} = \sqrt{\frac{E}{E'}}$

3. Squaring both sides,

$\frac{1}{4} = \frac{E}{E'}$

So,

$E' = 4E$

But

$E' = E + \Delta E$

Therefore,

$E + \Delta E = 4E$

$\Delta E = 3E$

4. Checking options:

• A: $E$ ❌
• B: $3E$ ✅
• C: $2E$ ❌
• D: $4E$ ❌

Therefore, the correct answer is B: $3E$.