In a mass spectrograph, an ion X of mass number 24 and charge +e and another ion Y of mass number 22 and charge +2 e enter in a perpendicular magnetic field with the same velocity. The ratio of the radii of the circular path in the field will be

Question

In a mass spectrograph, an ion X of mass number 24 and charge +e and another ion Y of mass number 22 and charge +2 e enter in a perpendicular magnetic field with the same velocity. The ratio of the radii of the circular path in the field will be (A) (11/22) (B) (11/2) (C) (22/11) (D) (24/11)

Tardigrade Admin · Accepted Answer

As, r=(m v/q B) i.e., r propto m / q So, (r1/r2)=(m1/m2) × (q2/q1) =(24 × 2 e/22 × e )=(24/11)

Q. In a mass spectrograph, an ion $X$ of mass number 24 and charge $+ e$ and another ion $Y$ of mass number 22 and charge $+ 2 e$ enter in a perpendicular magnetic field with the same velocity. The ratio of the radii of the circular path in the field will be

$\frac{11}{22}$

$\frac{11}{2}$

$\frac{22}{11}$

$\frac{24}{11}$

As, $r = \frac{m v}{qB}$ i.e., $r \propto m / q$
So, $\frac{r _{1}}{r _{2}} = \frac{m _{1}}{m _{2}} \times \frac{q _{2}}{q _{1}}$
$= \frac{24 \times 2 e}{22 \times e} = \frac{24}{11}$

Q. In a mass spectrograph, an ion X of mass number 24 and charge +e and another ion Y of mass number 22 and charge +2e enter in a perpendicular magnetic field with the same velocity. The ratio of the radii of the circular path in the field will be

2211​

211​

1122​

1124​