A particle moving with kinetic energy {E} has de Broglie wavelength
Question:

A particle moving with kinetic energy $\mathrm{E}$ has de Broglie wavelength $\lambda$. If energy $\Delta \mathrm{E}$ is added to its energy, the

wavelength become $\frac{\lambda}{2}$. Value of $\Delta \mathrm{E}$, is:

  1. (1) $\mathrm{E}$

  2. (2) $4 \mathrm{E}$

  3. (3) $3 \mathrm{E}$

  4. (4) $2 \mathrm{E}$


Correct Option: , 3

Solution:

(3) As per question, when $\mathrm{KE}$ of particle $E$, wavelength $\lambda$ and when $K E$ becomes $E+\Delta E$ wavelength becomes $\lambda / 2$

Using, $\lambda=\frac{h}{\sqrt{2 m K E}}$

$\frac{\lambda}{2}=\frac{h}{\sqrt{2 m(K E+\Delta E)}}$

$\Rightarrow \frac{\lambda}{/ 2}=\sqrt{\frac{K E+\Delta E}{K E}}$

$\Rightarrow 4=\frac{K E+\Delta E}{K E}$

$\Rightarrow \quad 4 K E-K E=\Delta E$

$\therefore \quad \Delta E=3 \underline{K E}=3 \underline{E}$

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