CUET Preparation Today
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$\text{Debroglie Wavelength } \lambda = \frac{h}{p} = \frac{h}{\sqrt {2mE}}$ $\lambda' = \frac{\lambda}{2} = \frac{h}{\sqrt {2m(E+\Delta E)}}$ $\frac{1}{2}\frac{h}{\sqrt {2mE}} = \frac{h}{\sqrt {2m(E+\Delta E)}}$ $\Rightarrow (E+\Delta E) = 4E $ $\Rightarrow \Delta E = 3E$ |
