Assume that the de Brogie wave associated with an electron can can from a standing wave between the atome arrange in a one dimensional array with nodes at each of the atomic sites it is found that one such standing wave if the distance d between the aloms of the arry is `2 M139` wave is again formad if d is increased to `2.5Å` . A similar standing in the distance if d find the energy of the electrons velts and tghe least value of d for which the standing wave type described above can from .

Answer

A

Explanation

As nodes are formed at each of the atomic sites , heance
`2Å= n((lambda)/(2)) `…(i)
[:'`Distance between successive nodes ` = lambda//2]`


`and 2.5 Å= (n + 1) (lambda)/(2)`
:. (2.5)/(2) = (n + 1)/(n) ,(5)/(4) = (n + 1)/(n) or n = 4 `
Heance , from equatio(1).
` 2.5 Å= 4 (lambda)/(2) i,e , lambda = 1Å`
Now , de broglie wavelength is given by
`lambda = (h)/(sqrt(2mK)) orK = (h^(2))/(lambda^(2)2m )`
`:. K = ((6.63 xx 10^(-34))^(2))/((1 xx 10^(-10) ) ^(2) xx 2 xx 9.1 xx 10^(-31) xx 1.6 xx 10^(-10))eV`
`= ((6.63))^(2))/(6 xx 9.1 xx 1.6) xx 10^(2) eV = 151 eV`
d will be miximum , when
`n = 1 , d_(min ) = (lambda)/(2) = (1Å)/(2) = 0.5Å`