The masses of two isotopes of chlorine are 34.980 and 36.978. If the radius of the circular path in Bainbridge spectrograph corresponding to lighter is 5 cm, the distance between the spots on photographic plate marked by the two isotopes will be
A. 5.7 cm
B. 0.57 cm
C. 0.57 mm
D. 0.57 m

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Hint: Inside a spectrograph, particles move at a particular speed in a plane perpendicular to a magnetic field. Due to the action of the magnetic field the articles will move in a circle, we can use the Lorentz force formula to find the radius in which those particles move.

Formula used:
Lorentz force
F = q(v×B)

Complete step by step answer:
The Lorentz force will provide the necessary centripetal acceleration inside the spectrograph. Both the isotopes will have the same amount of charge and velocity on them. And as the direction of motion of particles and magnetic fields are perpendicular to each other, their cross product will be equal to their arithmetic product.
  & \dfrac{m{{v}^{2}}}{r}=qvB \\
 & \Rightarrow r=\dfrac{mv}{qB} \\
So, the radius will be proportional to the mass of the particle in this case as all other quantities are the same for both the particles.
  & \dfrac{r}{5}=\dfrac{36.978}{34.980} \\
 & \Rightarrow r=5.285 \\
As we can see the radius of the circular path in the Bainbridge spectrograph corresponding to heavier isotope will be 5.285 cm.

As we can see in the image, the distance between the two points on the photographic plate will be twice the difference of radii of the two circles. The difference between the radii is 0.285 cm and twice that will be 0.57 cm.

Hence, the correct option is B, i.e. 0.57 cm.

There can be a common mistake of not taking twice the difference of radii if the setup of the Bainbridge spectrograph is not visualized. The distance between the starting point and the point at which the particle strikes the plate is twice the radius made by the particle and both particles start at the same point.