Assume that the de Broglie wave associated with an electron can from a standing wave between the atom 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 atoms of the array is `2 A^0` A similar standing wave is again formed if d is increased to `2.5Å` . Find the energy of the electrons in electron volts and the least value of d for which the standing wave type described above can from .

Options

A. `0.4 A^(@)`
B. `0.5A^(@)`
C. `2A^(@)`
D. `1A^(@)`

Answer

B

Explanation

From the figure it is clear that
`p.(lambda//2)= 2Å`
`(p+1)lambda//2=2.5Å`

`:. Or lambda//2=(2.5-2.0)Å=0.5Å`
or `lambda=1Å=10^(-10)m`
(i) de Broglie wavelength is given by
`lambda(h)/(p)=(h)/sqrt(2Km)` K = Kinetic energy of electron
`:. K=(h)/(2mlambda^(2))`
`((6.63xx10^(-34))^(2))/(2(9.1xx10^(-31))(10^(-10))^(2))=2.415xx10^(-17)J`
`=((2.145xx10^(-17))/(1.6xx10^(-19)))eV`
`:. K=150.8 eV`
(ii)
The least value of `d` will be when only one loop is formed
`d_(min)=lambda//2`
`:. d_(min)=lambda//2 or d_(min)=0.5Å`