A particle moving with kinetic energy E has de Broglie wavelength . If energy is added to its energy, the wavelength become . Value of , is

JEE MainPhysicsModerateModern physics

A particle moving with kinetic energy E has de Broglie
wavelength $\lambda$ . If energy $\Delta E$ is added to its energy, the
wavelength become $\lambda/2$ . Value of $\Delta E$ , is

Show answer and solution

\lambda = \frac{h}{\sqrt{2(KE)m}} = \lambda \propto \frac{1}{\sqrt{KE}}

\frac{\lambda}{\lambda/2} = \sqrt{\frac{{KE}_{f}}{{KE}_{i}}}

4{KE}_{i} = {KE}_{f}

\Rightarrow \Delta E = 4KE_{i} - KE_{i} = 3KE = 3E

Source: Grass-G / P-19 Physics paper 28 Nov Fourth.docx

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