Let us assume that the de-Broglie wave asssociated with an electron froms a standing wave between the atoms arranged in a one-dimensional array with nodes at each of the atomic sites. If is found that one such standing wave is formed, if the distance between the atoms of the array is `3 Å`. A similar standing wave is again formed, if d is increased to `3.5 Å` but not for any intermediate value of d. Find the energy of the electrons in electron volts and the least value of d for which the standing wave of the type described above can from.

Answer

(i) `150.8eV` (ii) `0.5Å`

Explanation

From the figure it is clear that
`(P+1). Lambda//2=2.5 Å`
`:. Lambda//2=(2.5-2.0)Å=0.5 Å`
or `lambda=1 Å=10^(-10) m`.
de-Broglie wavelength is given by
`lambda=h/p=(h)/sqrt(2Km)`
K= kinetic energy of electron
`:. K=h^(2)/(2mlambda^(2))=((6.63xx10^(-34))^(2))/(2(9.1xx10^(-31))(10^(-10))^(2))=2.415xx10^(-17) J`
`=((2.145xx10^(-17))/(1.6xx10^(-19)))eV=150.8 eV`