MCQ
A particle moving with kinetic energy $E$ 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
- A$2E$
- B$E$
- ✓$3E$
- D$4E$
✓
Answer
Correct option: C.
$3E$
c
de-Broglie wavelength $=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}=\lambda$
de-Broglie wavelength $=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}=\lambda$
Also, $\frac{\mathrm{h}}{\sqrt{2 \mathrm{m}(\mathrm{E}+\Delta \mathrm{E})}}=\frac{\lambda}{2}$
$\therefore \quad \frac{\mathrm{E}+\Delta \mathrm{E}}{\mathrm{E}}=4 \Rightarrow \Delta \mathrm{E}=3 \mathrm{E}$
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