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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 on each of the atomic site. It is found that one such standing wave is formed if the distance between the atoms of the array is $2\: A^{\circ}$ A similar standing wave is again formed if $d$ is increased to $2.5\: A^{\circ}$ . but not for any other intermediate value of $d$. The energy of the electron is

$\begin {array} {1 1} (a)\;15.095 \: eV & \quad (b)\;1509.5\: eV \\ (c)\;150.095\: eV & \quad (d)\;150.95\: eV \end {array}$

 

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Ans : (d)
Interatomic spacing depends upon the path difference. Interatomic spacing is
$d =\large\frac{n\lambda}{2}$ and de- Broglie wave is formed between interatomic spacing.
$d_1 = \large\frac{n \lambda}{2} \: and\: d_2 = \large\frac{(n+1) \lambda}{2}$
So,$ \large\frac{ \lambda}{2} = d_2 – d_1$
$ \Rightarrow \lambda = 2(d_2 – d_1) = 2(2.5 – 2) = 1\:  A^{\circ}$
$ \lambda=\large\frac{h}{p} \: \: or \: \: p = \large\frac{h}{ \lambda}$
$E = \large\frac{1}{2} mv^2$
$= \large\frac{1}{2m} (m^2v^2 ) = \large\frac{p^2}{2m}$
$ \Rightarrow E = \large\frac{h^2}{2m \lambda^2}$
$ \Rightarrow E = \large\frac{(6.6 \times 10^{-34})^2}{ [ 2 \times 9.1 \times 10^{-31} \times (10^{-10} )^2 ]} = 150.95 \: eV$

 

answered Jan 1, 2014 by thanvigandhi_1
edited Nov 7, 2014 by thagee.vedartham
 

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