Given:
- The particles have equal charges \( q \).
- They are accelerated through the same potential difference \( V \).
- Radii of circular paths in a magnetic field are \( R_1 \) for particle \( X \) and \( R_2 \) for particle \( Y \).
Step 1. Relate radius to mass and velocity:
For a particle moving in a circular path in a magnetic field, the radius \( R \) is given by:
\(R = \frac{mv}{qB}\)
where:
- \( m \) is the mass of the particle,
- \( v \) is the velocity after acceleration,
- \( q \) is the charge, and
- \( B \) is the magnetic field strength.
Step 2. Express \( v \) in terms of \( V \):
Since each particle is accelerated through the same potential difference \( V \), the kinetic energy gained by each particle is:
\(\frac{1}{2} mv^2 = qV\)
Solving for \( v \), we get:
\(v = \sqrt{\frac{2qV}{m}}\)
Step 3. Substitute \( v \) into the radius formula:
\( R = \frac{m}{qB} \cdot \sqrt{\frac{2qV}{m}} = \frac{\sqrt{2m \cdot qV}}{qB}\)
Step 4. Determine the ratio of radii for particles \( X \) and \( Y \):
\(\frac{R_1}{R_2} = \frac{\sqrt{2m_X \cdot qV}}{\sqrt{2m_Y \cdot qV}} = \sqrt{\frac{m_X}{m_Y}}\)
Step 5. Solve for the mass ratio:
Squaring both sides, we get:
\(\frac{m_X}{m_Y} = \left(\frac{R_1}{R_2}\right)^2\)
Thus, the mass ratio \( \frac{m_X}{m_Y} \) is \( \left(\frac{R_1}{R_2}\right)^2 \).
The Correct Answer is: \( \left( \frac{R_1}{R_2} \right)^2 \)
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: