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Question

(C) The de Broglie wavelength $\lambda$ of a particle with kinetic energy $E$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$.When the energy is increase

Vedclass · Accepted Answer

$3E$

Vedclass · Answer

(C) The de Broglie wavelength $\lambda$ of a particle with kinetic energy $E$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$.When the energy is increased by $\Delta E$,the new kinetic energy becomes $E' = E + \Delta E$,and the new wavelength is $\lambda' = \frac{\lambda}{2}$.Using the formula for the new wavelength: $\lambda' = \frac{h}{\sqrt{2m(E + \Delta E)}}$.Since $\lambda' = \frac{\lambda}{2}$,we have $\frac{h}{\sqrt{2m(E + \Delta E)}} = \frac{1}{2} \cdot \frac{h}{\sqrt{2mE}}$.Squaring both sides: $\frac{1}{2m(E + \Delta E)} = \frac{1}{4} \cdot \frac{1}{2mE}$.Simplifying the equation: $\frac{1}{E + \Delta E} = \frac{1}{4E}$.Therefore,$4E = E + \Delta E$,which gives $\Delta E = 3E$.

$A$ particle moving with kinetic energy $E$ has de Broglie wavelength $\lambda$. If energy $\Delta E$ is added to its energy,the wavelength becomes $\frac{\lambda}{2}$. The value of $\Delta E$ is:

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