Question

An experiment was carried out twice using magnetic deflection type mass-spectrometer to estimate a certain molecule present in the compound under test. In the first test, the sample was excited and the potential difference between the accelerating plate was kept at 1000 V and in the second case, the potential difference was at 10000 V. Determine the ratio of the radius of curvature of the circular path travelled by the ions when they pass through the magnetic field oriented right angle to their motion in the test to that of the second

• A. $1/\sqrt{10}$
• B. $1/10$
• C. $\sqrt{10}$
• D. $10$

Hint:

Remember the direct proportionality $R \propto \sqrt{V}$ in magnetic deflection mass spectrometers.
If the voltage increases by a factor of $k$, the radius of the path increases by a factor of $\sqrt{k}$.
This allows rapid computation without recalling the entire derivation.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks for the ratio of the radii of curvature of the circular paths of ions in a magnetic deflection mass spectrometer under two different accelerating voltages ($V_1 = 1000\text{ V}$ and $V_2 = 10000\text{ V}$).
The ions have the same mass and charge, and they enter the same magnetic field.

Step 2: Key Formula or Approach:
An ion of charge $q$ and mass $m$ accelerated through a potential difference $V$ gains kinetic energy:
\[ qV = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2qV}{m}} \] When the ion enters a perpendicular magnetic field $B$, the magnetic force provides the necessary centripetal force:
\[ qvB = \frac{mv^2}{R} \implies R = \frac{mv}{qB} \] Substituting the velocity $v$ into the radius formula:
\[ R = \frac{m}{qB}\sqrt{\frac{2qV}{m}} = \frac{1}{B}\sqrt{\frac{2mV}{q}} \] This shows that the radius of curvature is directly proportional to the square root of the accelerating voltage, $R \propto \sqrt{V}$.

Step 3: Detailed Explanation:

• Let $R_1$ be the radius of curvature at voltage $V_1 = 1000\text{ V}$.

• Let $R_2$ be the radius of curvature at voltage $V_2 = 10000\text{ V}$.

• Since $R \propto \sqrt{V}$, we can express the ratio as:
\[ \frac{R_1}{R_2} = \sqrt{\frac{V_1}{V_2}} \]
• Substituting the given values:
\[ \frac{R_1}{R_2} = \sqrt{\frac{1000}{10000}} = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}} \]

Step 4: Final Answer:
The ratio of the radius of curvature of the circular path in the first test to that of the second is $1/\sqrt{10}$.