Charge = q = $1.6 times 10^{-19} C$

q = $1.6 times 10^{-19} C$

$$
frac{E}{B}=frac{frac{textrm{N}}{textrm{C}}}{frac{textrm{N}}{textrm{A}cdot textrm{m}}}
$$

$$
frac{E}{B}=frac{textrm{A}cdottextrm{m}}{textrm{C}}=frac{textrm{C}cdottextrm{m}}{textrm{C}cdottextrm{s}}
$$

Which finally gives

$$
boxed{frac{E}{B}=frac{textrm{m}}{textrm{s}}}
$$

Q.E.D.

b) X-rays

c) All have the same speed, $c$.

$$
E=vB
$$

From which we express the magnetic field as
$B=frac{E}{v}=frac{5.8times 10^3}{3.6times 10^4}$
Finally,

$$
boxed{B=0.16textrm{ T}}
$$

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

From here, we can express the radius to have that

$$
r=frac{mv}{qB}=frac{1.67times 10^{-27}times 5.4times 10^4}{1.6times 10^{-19}times 0.06}
$$

Finally, we have that

$$
boxed{r=9.4times 10^{-3}textrm{ m}}
$$

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

However, we have to find the speed first. We can do this by employing the conservation of energy which says

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

So we get that

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

Now, we obtain that

$$
qB=frac{m}{r}cdotsqrt{frac{2qV}{m}}
$$

which gives that the magnetic field is given as

$$
B=sqrt{frac{2mV}{qr^2}}=sqrt{frac{2times9.11times 10^{-31}times 4.5times 10^3}{1.6times 10^{-19}times 5^2times 10^{-4}}}
$$

Which finally gives that

$$
boxed{B=4.5times 10^{-3}textrm{ T}}
$$

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

From which mass can be expressed as

$$
m=frac{qr^2B^2}{2V}=frac{2times 1.6times 10^{-19}times 8^2times 10^{-4}times 0.077^2}{2times 156}
$$

Finally, one gets that

$$
boxed{m=3.9times 10^{-26}textrm{ kg}}
$$

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

a) From here, one can express the speed as

$$
v=frac{qrB}{m}=frac{2times 2times 1.6times 10^{-19}times 0.15}{6.6times 10^{-27}}
$$

Finally, we have that

$$
boxed{v=1.45times 10^7frac{textrm{ m}}{textrm{ s}}}
$$

b) The kinetic energy is to be found via its standard expression

$$
K=frac{mv^2}{2}=frac{6.6times 10^{-27}times 1.45^2times 10^{14}}{2}
$$

Which gives that

$$
boxed{K=6.9times 10^{-13}textrm{ J}}
$$

c) Now, the potential difference that corresponds to that energy is found from

$$
K=qV
$$

$$
V=frac{6.9times 10^{-13}}{2times1.6times 10^{-19}}
$$

Eventually, we get

$$
boxed{V=2.2 times 10^6textrm{ V}}
$$

$$
textrm{b) }K=6.9times 10^{-13}textrm{ J}
$$

$$
textrm{c) }V=2.2 times 10^6textrm{ V}
$$

$$
S=(1-frac{175times10^3}{175times10^3+1})times 100
$$

Which gives that

$$
boxed{S=0.0006%}
$$

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

Here we see that $mpropto r^2$ so we can write

$$
frac{m_1}{m_2}=frac{r_1^2}{r_2^2}
$$

which gives us that the mass of the second isotope is

$$
m_2=m_1times frac{r_2^2}{r_1^2}=28m_ptimes frac{0.1797^2}{0.1623^2}
$$

Which finally gives that

$$
boxed{m_2=34m_p}
$$

$$
lambda=frac{c}{f}=frac{3times 10^8}{66times 10^6}
$$

Which gives that

$$
lambda=4.5textrm{ m}
$$

Now, the distance between the bars of the antenna is

$$
d=frac{lambda}{4}=frac{4.5}{4}
$$

Finally,

$$
boxed{d=1.1textrm{ m}}
$$

$$
f=frac{c}{lambda}=frac{3times 10^8}{650times 10^{-9}}
$$

Which gives that

$$
boxed{f=4.6times 10^{14}textrm{ Hz}}
$$

$$
lambda=frac{c}{f}=frac{3times 10^8}{101.3times 10^6}
$$

Which gives that

$$
lambda=2.96textrm{ m}
$$

Now, the optimum length of the antenna is $d=frac{lambda}{2}$

$$
d=frac{lambda}{2}=frac{2.96}{2}
$$

Finally,

$$
boxed{textrm{d=1.48textrm{ m}}}
$$

$$
lambda=frac{c}{f}=frac{3times 10^8}{8times 10^8}
$$

Which gives that

$$
lambda=0.375textrm{ m}
$$

Now, the optimum length of single-end antennas is $d=frac{lambda}{4}$ as given in the problem

$$
d=frac{lambda}{4}=frac{0.375}{4}
$$

Finally,

$$
boxed{d=0.0938textrm{ m}}
$$

$$
N=frac{2.7times 10^{-26}}{1.67times 10^{-27}}
$$

$$
boxed{N=16textrm{ amu}}
$$

$$
lambda=frac{c}{f}=frac{3times 10^8}{94.5times 10^6}
$$

Which gives that

$$
lambda=3.17textrm{ m}
$$

Now, the optimum length of an antenna is $d=frac{lambda}{2}$

$$
d=frac{lambda}{2}=frac{3.17}{2}
$$

Finally,

$$
boxed{d=1.59textrm{ m}}
$$

$$
d=frac{lambda}{4}
$$

So we can express the wavelength as

$$
lambda=4d
$$

On the other hand, the frequency is given as

$$
f=frac{c}{lambda}=frac{c}{4d}=frac{3times 10^8}{4times 0.083}
$$

Finally, we have that

$$
boxed{f=9.04times 10^8textrm{ Hz}}
$$

$$
frac{q}{m}=frac{2V}{r^2B^2}=frac{2times 150}{0.098^2times50^2 times 10^{-6}}
$$

Finally, the charge-to-mass ratio is

$$
boxed{frac{q}{m}=12.5times 10^{6}frac{textrm{ C}}{textrm{ kg}}}
$$

$$
v_{target}=frac{cf_{Doppler}}{2f_{transmitted}}=frac{3times 10^8times 1850}{2times 10.525times 10^9}
$$

Which finally gives that

$$
boxed{v=26.4frac{textrm{m}}{textrm{s}}}
$$

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

So we can express the radius as

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

Since we know that our detector has the spacing $d=0.1times 10^{-3}$ m we have that

$$
d=2times(r_{176}-r_{175})=2times(sqrt{frac{2m_{176}V}{qB^2}}-sqrt{frac{2m_{175}V}{qB^2}})
$$

Which after we transfer the masses into proton masses gives

$$
d=2timessqrt{frac{2m_pV}{qB^2}}(sqrt{176}-sqrt{175})
$$

Therefore, if we take that $V_{min}=500$ V we can express $B$ as

$$
B=2timesfrac{sqrt{2Vm_p}(sqrt{176}-sqrt{175})}{dsqrt{q}}=2timesfrac{sqrt{2times 500times1.67times 10^{-27}}(sqrt{176}-sqrt{175})}{0.1times 10^{-3}times sqrt{1.6times 10^{-19}}}
$$

Finally, the magnetic field should be

$$
boxed{B=2.4textrm{ T}}
$$

$$
a=frac{L}{y_1}lambda
$$

Where $L$ is the distance to the screen. After we plug in the values

$$
a=frac{0.95}{8.5times 10^{-3}}times 633times 10^{-9}
$$

Finally, the width of the slit is

$$
boxed{a=71times 10^{-6}textrm{ m}}
$$

$F = dfrac{K (4q) (2q)}{r^2}$

$F = dfrac{8 K q^2}{r^2}$

The total charge on the spheres is:

$q_{net} = q_1 + q_2 = (-4q) + (2q) = -2q$

After connecting the spheres, charges on each of the spheres is the half of the $q_{net}$:

$q’_1 = q’_2 = -q$

The force between them:

$F’ = dfrac{K q_1 q_2}{r^2}$

$= dfrac{K q^2}{r^2}$

$$
= dfrac{1}{8} F
$$