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Assume that the de-Broglie wave associated with an electron can form a standing wave between the atoms arranged in a one-dimensional array with nodes at each of the atomic sites. It is found that one such standing wave is formed if the distance $d$ between the atoms of the array is $2 Å$ . A similar standing wave is again formed if $d$ is increased to $2.5 Å$ but not for any intermediate value of $d$ . Find the energy of the electron in $e V$ and the least value of $d$ for which the standing wave of the type described above can form.

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$501.3 eV, 3.5 Å$

$309.5 eV, 1.5 Å$

$150.9 eV, 0.5 Å$

$403.2 eV, 2.5 Å$

answer is C.

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Detailed Solution

From the figure, it is clear the number of half wavelengths differ by one.

$So, p \cdot (\frac{λ}{2}) = 2 Å and$

$(p + 1) \cdot \frac{λ}{2} = 2.5 Å$

$\Rightarrow \frac{λ}{2} = (2.5 - 2) Å = 0.5 Å$

$\Rightarrow λ = 1 Å = 10^{- 10} m$

de-Broglie wavelength is given by

$λ = \frac{h}{p} = \frac{h}{\sqrt{2 m K}}$

where $K$ is the kinetic energy of electron

$\Rightarrow K = \frac{h^{2}}{2 m λ^{2}}$

$\Rightarrow K = \frac{{(6.63 \times 10^{- 34})}^{2}}{2 (9.1 \times 10^{- 31}) {(10^{- 10})}^{2}}$

$\Rightarrow K = 2.415 \times 10^{- 17} J$

$\Rightarrow K = (\frac{2.415 \times 10^{- 17}}{1.6 \times 10^{- 19}}) e V$

$\Rightarrow K = 150.9 e V$

The least value of $d$ will be, when only one loop is formed. So, we have

$d_{\min} = \frac{λ}{2}$

$\Rightarrow d_{\min} = 0.5 Å$

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