In a mass spectrometer chlorine ions of mass 35u and charge +5e are emitted from a source and accelerated through a potential difference of 250 kV. They then enter a region with a magnetic field which is perpendicular to their original direction of motion. The chlorine ions exit the spectrometer after being bent along a path with radius of curvature 3.5 m. What is the value of the magnetic field? (u = 1.66

Respuesta :

Answer:

Magnetic field will be equal to 0.269 T

Explanation:

We have given mass of chlorine ion is 35u

As we know that 1 u = [tex]=1.66\times 10^{-27}kg[/tex]

Radius of circular path is given r = 3.5 m

So mass of chlorine ion [tex]=35\times 1.66\times 10^{-27}kg=58.1\times 10^{-27}kg[/tex]

Charge [tex]q=1.6\times 10^{-19}C[/tex]

Potential difference V = 250 KV

From conservation of energy

[tex]\frac{1}{2}mv^2=qV[/tex]

So [tex]\frac{1}{2}\times 58.1\times 10^{-27}\times v^2=1.6\times 10^{-19}\times 250000[/tex]

[tex]v=2.6\times 10^6m/sec[/tex]

We know that radius is equal to [tex]r=\frac{mv}{qB}[/tex]

So [tex]3.5=\frac{58.1\times 10^{-27}\times 2.6\times 10^6}{1.6\times 10^{-19}\times B}[/tex]

[tex]B=0.269Tesla[/tex]