Kinetic energy of the particle is E and it's de-Broglie wavelength is $λ$. On increasing its K.E by ΔE, it's new de-Broglie wavelength becomes $\frac{λ}{2}$. Then ΔE is
Answer & explanation
Correct answer: option 1
$\text{Debroglie Wavelength } \lambda = \frac{h}{p} = \frac{h}{\sqrt {2mE}}$
$\lambda' = \frac{\lambda}{2} = \frac{h}{\sqrt {2m(E+\Delta E)}}$
$\frac{1}{2}\frac{h}{\sqrt {2mE}} = \frac{h}{\sqrt {2m(E+\Delta E)}}$
$\Rightarrow (E+\Delta E) = 4E $
$\Rightarrow \Delta E = 3E$