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# TOF-Chemistry AQA watch

1. The image is too small for my poor old eyes ...
2. (Original post by charco)
The image is too small for my poor old eyes ...
sorry for that

https://imgur.com/a/819l0vW
3. d is a constant and

the kinetic energy is the same for all particles

subscript '1' refers to the 82- isotope

Hence, 1/2m1(v1)2 = 1/2m2(v2)2

m1(v1)2 = m2(v2)2

m1/m2 = (v2)2/(v1)2

v can be substituted for d/t (for proportionality the d's are constant and can be omitted)

and the mass ratio is 82/86 = 0.953

0.953 is proportional to t12/t22

t1 = 1.243 x 10-5

t22 = 1.545 x 10-10/0.953 = 1.621 x 10-10

Hence t2 = 1.273 x 10-5

Not really a chemistry question ...

Better check my working out. You would expect the heavier ion to travel slower, so it's not an unlikely answer.
4. (Original post by charco)
d is a constant and

the kinetic energy is the same for all particles

subscript '1' refers to the 82- isotope

Hence, 1/2m1(v1)2 = 1/2m2(v2)2

m1(v1)2 = m2(v2)2

m1/m2 = (v2)2/(v1)2

v can be substituted for d/t (for proportionality the d's are constant and can be omitted)

and the mass ratio is 82/86 = 0.953

0.953 is proportional to t12/t22

t1 = 1.243 x 10-5

t22 = 1.545 x 10-10/0.953 = 1.621 x 10-10

Hence t2 = 1.273 x 10-5

Not really a chemistry question ...

Better check my working out. You would expect the heavier ion to travel slower, so it's not an unlikely answer.
Iirc from last year this is definitely correct
5. (Original post by charco)
d is a constant and

the kinetic energy is the same for all particles

subscript '1' refers to the 82- isotope

Hence, 1/2m1(v1)2 = 1/2m2(v2)2

m1(v1)2 = m2(v2)2

m1/m2 = (v2)2/(v1)2

v can be substituted for d/t (for proportionality the d's are constant and can be omitted)

and the mass ratio is 82/86 = 0.953

0.953 is proportional to t12/t22

t1 = 1.243 x 10-5

t22 = 1.545 x 10-10/0.953 = 1.621 x 10-10

Hence t2 = 1.273 x 10-5

Not really a chemistry question ...

Better check my working out. You would expect the heavier ion to travel slower, so it's not an unlikely answer.

thanks

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