Question

g What are the relative rates of diffusion of the three naturally occurring isotopes of krypton, 80Kr80Kr , 82Kr82Kr, and 83Kr83Kr. What are the relative rates of diffusion of the three naturally occurring isotopes of krypton, , , and . The relative rates of diffusion are: 82Kr(1.02)>82Kr(1.02)>83Kr(1.01)>80Kr(1.00)83Kr(1.01)>80Kr(1.00) The relative rates of diffusion are: 80Kr(1.02)>80Kr(1.02)>82Kr(1.01)>83Kr(1.00)82Kr(1.01)>83Kr(1.00) The relative rates of diffusion are: 82Kr(1.02)>82Kr(1.02)>80Kr(1.01)>83Kr(1.00)80Kr(1.01)>83Kr(1.00) The relative rates of diffusion are: 83Kr(1.02)>83Kr(1.02)>82Kr(1.01)>80Kr(1.00)82Kr(1.01)>80Kr(1.00)

Answer 1

According to the Graham's law of diffusion, we know that, the rate of the diffusion varies inversely to the molar mass of the gas, i.e.

Rate of diffusion,
`$r_d = (a)/(\sqrt M)$`

where, the 'M' is the molar mass of the gas.

Now in the case of the isotopes of the Krypton,

Atomic mass of
`$^(80)Kr$` = 80 AMU

Atomic mass of
`$^(82)Kr$` = 82 AMU

Atomic mass of
`$^(83)Kr$` = 83 AMU

So the ratio of the rate of diffusion of the three isotopes are :

`$M_(d,^(80)Kr):M_(d,^(82)Kr):M_(d,^(83)Kr)$`

`$=(1)/(√(80)):(1)/(√(82)):(1)/(√(83))$`

`$=0.1118:0.1104:0.10976$`

Dividing the above three with the smallest number among the three i.e. 0.10976, we get the relative rates of diffusion.

∴
`$M_(d,^(80)Kr):M_(d,^(82)Kr):M_(d,^(83)Kr)$`

= 1.02 : 1.01 : 1

Hence the relative rate of diffusion are :

`$^(80)Kr(1.02)>^(82)Kr(1.01)>^(83)Kr(1.00)$`