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(i) Explain the shape of s and p-orbital. (ii) The mass of an electron is 9.1xx10^(-31)kg. If its kinetic energy is 3.0xx10^(-25) J. |
Answer» Solution :(i) s-orbital: For Is orbital,l=0, m=0, `f(theta)=1//sqrt2` and `g(varphi)=1//sqrt2pi`.. Therefore, the angular distribution function is equal to `1//2sqrtpi`. i.e. it is INDEPENDENT of the angle 0 and Hence, the probability of finding the ELECTRON is independent of the direction from the nucleus. So, the shape of the s orbital is spherical. p-orbital: For p orbitals l=1 and the corresponding m values arc-1, 0 and +1.Th. three different m values indicates that there are three different orientations possible for orbitals. These orbitals are designated as `p_(s),p_(y)` and `p_(s)`. The shape of p orbitals are dum bell shape.  Step I. Calculate of the velocity of electron KINETIC energy `=1//2 mv^(2)=3.0xx10^(-25)J=3.0xx10^(-25) Kg m^(2)s^(-2)` `v^(2)=(2xxK.E)/(m)=(2xx(3.0xx10^(-25) Kg m^(2)s^(-2)))/(9.1xx10^(-31) kg)=65.9xx10^(4)m^(2) s^(-2)` `v=(65.9xx10^(4) ms^(-2))^(1//2)=8.12xx10^(2)ms^(-1)` Step II. Calculation ofwave LENGTH of theelectron AAccording to de Broglie.s equation , `lambda=(h)/(mv)=((6.626xx10^(-34)Kg m^(2)s^(-1)))/((9.1xx10^(-31) Kg)XX(8.12xx 10^(2) ms^(-1)))` `=0.08967xx10^(-5)m=8967xx10^(-10)m=8967A(:.1A=10^(-10)m)` |
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