Assume that the de-Broglie were 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 eV and the least value of d for which the standing wave of the type described above can form.

Explanation

For standing wave , the separation between consecutive nodes `("or antinodes")` is `lambda//2`
:. `d = (n lambda)/(2)`, where`n` is integer (i)
For next node
`D = (n + 1) (lambda)/(2)`
`:. d - d = (lambda)/(2)`
`implies lambda = 2 (d- d)`
Here `d = 2 Å,d' = 2.5Å`
`:. lambda = 2 (2.5 Å- 2Å) = 1 Å`
Therefore, energy of electrons,
`E = (p^(2))/(2 m)`
`implies = (h^(2))/(2 m lambda^(2))`
Here `h = 6.63 xx 10^(-34) J-s`
`m = 9.1 xx 10^(-31) kg`
`lambda = 1 Å = 10^(-31) m`
`:. E = ((6.63 xx 10^(-34))^(2))/(2 xx 9.1 xx 10^(-31) xx (10^(-10))^(2)) J`
`= ((6.63)/(2 xx 9.1) xx 10^(-17)` joule = 2.415 xx 10^(-17) J`
`= (2.415 xx 10^(-17))/(1.6 xx 10^(-19)) e V = 151 e V`
For least value of `d, n = 1` . Therefore, from Eq. (i)
`d_(min) = (lambda)/(2) = (1)/(2) Å = 0.5 Å`