In order to solve this problem we are going to use Newton’s second law which says that on a circular path it has to hold that

$$
qvB=frac{mv^2}{r}
$$

From here, we can express the radius to have that

$$
r=frac{mv}{qB}=frac{1.67times 10^{-27}times 7.5times 10^3}{1.6times 10^{-19}times 0.6}
$$

Finally, we have that

$$
boxed{r=1.3times 10^{-4}textrm{ m}}
$$

$$
r=1.3times 10^{-4}textrm{ m}
$$

In order to solve this problem we are going to use Newton’s third law which says that in equilibrium it has to hold that

$$
qvB=qE
$$

Which gives that the speed is given as

$$
v=frac{E}{B}
$$

So after we plug in the values

$$
v=frac{3000}{0.06}
$$

Finally

$$
boxed{v=50times 10^3frac{textrm{m}}{textrm{s}}}
$$

$$
v=50times 10^3frac{textrm{m}}{textrm{s}}
$$

In order to solve this problem, we are going to use Newton’s second law which says that on a circular path it has to hold that

$$
qvB=frac{mv^2}{r}
$$

However, we have to find the speed first. The speed is obtained while the electric field is acting on the electron
third law which says that in equilibrium it has to hold that

$$
qvB=qE
$$

Which gives that the speed is given as

$$
v=frac{E}{B}
$$

So after we plug in the values

$$
v=frac{3000}{0.06}=50times 10^3frac{textrm{m}}{textrm{s}}
$$

Now, we can express the radius to have that

$$
r=frac{mv}{qB}=frac{9.11times 10^{-31}times 50times 10^3}{1.6times 10^{-19}times 0.06}
$$

Finally, we have that

$$
boxed{r=4.7times 10^{-6}textrm{ m}}
$$

$$
r=4.7times 10^{-6}textrm{ m}
$$

In order to solve this problem we are going to use Newton’s third law which says that in equilibrium it has to hold that

$$
qvB=qE
$$

Which gives that the speed is given as

$$
v=frac{E}{B}
$$

So after we plug in the values

$$
v=frac{4500}{0.6}
$$

Finally

$$
boxed{v=7.5times 10^3frac{textrm{m}}{textrm{s}}}
$$

$$
v=7.5times 10^3frac{textrm{m}}{textrm{s}}
$$

In order to solve this problem, we are going to use Newton’s second law which says that on a circular path it has to hold that

$$
qvB=frac{mv^2}{r}
$$

However, we have to find the speed first. The final speed of the oxygen atoms is the consequence of the acceleration in the electric field which is expressed via the conservation of energy

$$
qV=frac{mv^2}{2}
$$

where we can express the square of the speed as

$$
v^2=frac{2qV}{m}
$$

Now, we will square our first equation and plug in the last expression into it

$$
q^2v^2B^2=frac{m^2v^4}{r^2}
$$

$$
q^2B^2=frac{m^2v^2}{r^2}=frac{m^2}{r^2}frac{2qV}{m}
$$

After some simple algebra one gets that

$$
qB^2=frac{2mV}{r^2}
$$

So we can express the mass as

$$
m=frac{qr^2B^2}{2V}=frac{1.6times 10^{-19}times 0.085^2times 0.072^2}{2times 110}
$$

Finally, we have that

$$
boxed{m=2.7times 10^{-26}textrm{ kg}}
$$

$$
m=2.7times 10^{-26}textrm{ kg}
$$

In order to solve this problem, we are going to use Newton’s second law which says that on a circular path it has to hold that

$$
qvB=frac{mv^2}{r}
$$

However, we have to find the speed first. The final speed of the argon atoms is the consequence of the acceleration in the electric field which is expressed via the conservation of energy

$$
qV=frac{mv^2}{2}
$$

where we can express the square of the speed as

$$
v^2=frac{2qV}{m}
$$

Now, we will square our first equation and plug in the last expression into it

$$
q^2v^2B^2=frac{m^2v^4}{r^2}
$$

$$
q^2B^2=frac{m^2v^2}{r^2}=frac{m^2}{r^2}frac{2qV}{m}
$$

After some simple algebra one gets that

$$
qB^2=frac{2mV}{r^2}
$$

So we can express the mass as

$$
m=frac{qr^2B^2}{2V}=frac{2times1.6times 10^{-19}times 0.106^2times 0.05^2}{2times 66}
$$

Finally, we have that

$$
boxed{m=6.8times 10^{-26}textrm{ kg}}
$$

$$
m=6.8times 10^{-26}textrm{ kg}
$$

In order to solve this problem we are going to use Newton’s third law which says that in equilibrium it has to hold that

$$
qvB=qE
$$

Which gives that the speed is given as

$$
v=frac{E}{B}
$$

So after we plug in the values

$$
v=frac{600}{0.0015}
$$

Finally

$$
boxed{v=400times 10^3frac{textrm{m}}{textrm{s}}}
$$

$$
v=400times 10^3frac{textrm{m}}{textrm{s}}
$$

In order to solve this problem, one has to understand that apart from the mass of the atoms, other parameters are the same and this results in different radii. From the charge-to-mass ratio, we have that

$$
frac{q}{m}=frac{2V}{r^2B^2}
$$

So we have that

$$
rpropto sqrt{m}
$$

Now, the ratio between the two isotopes is

$$
frac{r_{22}}{r_{20}}=sqrt{frac{m_{22}}{m_{20}}}
$$

We can express then the radii of the heavier isotope

$$
r_{22}=sqrt{frac{m_{22}}{m_{20}}}r_{20}=sqrt{frac{22}{20}}times 0.053
$$

Finally, we have that

$$
r_{22}=0.056textrm{ m}
$$

Now, the distance between the landing points of the two isotopes is equal to

$$
d=2(r_{22}-r_{20})=2times(0.056-0.053)
$$

So at the end we obtain

$$
boxed{d=0.006textrm{ m}}
$$

$$
d=0.006textrm{ m}
$$