Respuesta :
Answer: The minimum value of the potential difference through which these ions must be accelerated is [tex]1.87 \times 10^{-5}[/tex] MV.
Explanation:
The given data is as follows.
[tex]m_{1}[/tex] = 35 amu
= [tex]35 \times 1.66 \times 10^{-27}[/tex]
= kg
= 37 amu
= [tex]37 \times 1.66 \times 10^{-27}[/tex]
= kg
It is known that,
work done on charged particle = gain in kinetic energy
So,
v = [tex]\sqrt{(2 \times q \times \Delta_{V/m})}[/tex]
A centripetal force is experienced when charged particles enter a uniform magnetic field.
Hence, F = [tex]q \times v \times B (sin(90))[/tex]
[tex]\frac{m \times \frac{v^{2}}{r} = q \times v \times B[/tex]
r = [tex]\frac{m \times v}{B \times q}[/tex]
And,
[tex]r_{2} = \frac{m_{2} \times v_{2}}{(B \times q)}[/tex]
On completion of semi circle, the distance between two paths =
[tex]0.7 \times 10^{-2} = (\sqrt(2 \times m_{2} \times delta_{V/q}) -\sqrt{(2 \times m_{1} \times \frac{\Delta_{V/q}}{B}}[/tex]
[tex]0.7 \times 10^{-2} = \sqrt{(\Delta_{V})} \times (\sqrt(2 \times \frac{m_{2}}{q}) - \sqrt{\frac{(2 \times \frac{m_{1}}{q}))}{B}}[/tex]
[tex]\sqrt{(\Delta_V)} = 0.7 \times 10^{-2} \times \frac{B}{(\sqrt{(2 \times \frac{m_{2}{q})} - \sqrt{(\frac{2 \times m_{1}}{q}))}[/tex]
= [tex]\frac{0.7 \times 10^{-2} \times 1.2}{(\sqrt{(2 \times \frac{6.142 \times 10^{-26}}{(1.6 \times 10^{-19}})}) - \sqrt{(2 \times \frac{5.81 \times 10^{-26}}{(1.6 \times 10^{-19}))}[/tex]
= 350 volts
[tex]\Delta_V[/tex] = [tex]\sqrt{350}[/tex]
= 18.7 volts
= [tex]1.87 \times 10^{-5}[/tex] MV
Thus, we can conclude that the minimum value of the potential difference through which these ions must be accelerated is [tex]1.87 \times 10^{-5}[/tex] MV.
