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Step-by-Step Solution
Step 1: Write down the known information
A particle of mass $m$ has an initial kinetic energy $E$ and a corresponding de Broglie wavelength $\lambda$. When an additional energy $\Delta E$ is added, its wavelength becomes $\frac{\lambda}{2}$. Symbolically:
Initial kinetic energy: $E$
Initial wavelength: $\lambda = \frac{h}{\sqrt{2mE}}$
After adding $\Delta E$, the new energy: $E + \Delta E$
New wavelength: $\frac{\lambda}{2}$
Step 2: Express the new wavelength using de Broglie relation
According to the de Broglie relation, the wavelength is given by
$$\lambda' = \frac{h}{\sqrt{2m(E+\Delta E)}}.$$
Since $\lambda' = \frac{\lambda}{2}$, we set up the equation:
$$\frac{h}{\sqrt{2m(E+\Delta E)}} = \frac{\lambda}{2}.$$
Step 3: Substitute the initial wavelength $\lambda$ in the expression
We know
$$\lambda = \frac{h}{\sqrt{2mE}}.$$
So,
$$\frac{h}{\sqrt{2m(E+\Delta E)}} = \frac{1}{2}\,\frac{h}{\sqrt{2mE}}.$$
Step 4: Simplify the equation
Canceling $h$ and the factor $\sqrt{2m}$ on both sides, we get:
$$\frac{1}{\sqrt{E+\Delta E}} = \frac{1}{2}\,\frac{1}{\sqrt{E}}.$$
Rearranging,
$$\sqrt{E + \Delta E} = 2\,\sqrt{E}.$$
Step 5: Solve for $\Delta E$
Squaring both sides,
$$E + \Delta E = 4E.$$
Hence,
$$\Delta E = 4E - E = 3E.$$
Step 6: State the final result
The additional energy $\Delta E$ that needs to be added to make the de Broglie wavelength half of its original value is:
$$\Delta E = 3E.$$
