A nucleus X of mass M. intially at rest undergoes alpha decay according to the equation
`_(Z)^(A)Xrarr._(Z-2)^(A-4)Y+._(2)^(4)He`
The alpha particle emitted in the above proces is found to move in a circular track of radius r in a uniform magnetic field B. Then (mass and charge of `alpha` particle are m and q respectively)

Options

A. The ratio of kinetic energies of the alpha particle and (doughether nucleus Y approximately equals `(M-m)/(m)`
B. the ratio of kinetic energies of the alpha particle and the doughter nucleus Y apporximately `(m)/(M)`
C. The enetgy relased in the process approximately equals `(M)/(M-m)(r^(2)q^(2)B^(2))/(2m)`
D. The energy released in the process apporximately equals `(M)/(M-m)((r^(2)q^(2)B^(2))/(2m))`

Answer

To solve the problem of the alpha decay of nucleus X and the motion of the emitted alpha particle in a magnetic field, we can follow these steps:

### Step 1: Understand the decay process
The alpha decay process can be represented as:
\[ _{Z}^{A}X \rightarrow _{Z-2}^{A-4}Y + _{2}^{4}He \]
Here, nucleus X emits an alpha particle (which is a helium nucleus) and transforms into nucleus Y.

### Step 2: Identify the forces acting on the alpha particle
When the alpha particle moves in a magnetic field, it experiences a magnetic force that acts as the centripetal force required for circular motion. The magnetic force \( F_B \) is given by:
\[ F_B = qvB \]
where \( q \) is the charge of the alpha particle, \( v \) is its velocity, and \( B \) is the magnetic field strength.

### Step 3: Set up the centripetal force equation
The centripetal force required to keep the alpha particle moving in a circular path is given by:
\[ F_C = \frac{mv^2}{r} \]
where \( m \) is the mass of the alpha particle and \( r \) is the radius of the circular path.

### Step 4: Equate the forces
Since the magnetic force provides the centripetal force, we can set the two equations equal to each other:
\[ qvB = \frac{mv^2}{r} \]

### Step 5: Solve for velocity \( v \)
Rearranging the equation to solve for \( v \):
\[ v = \frac{qBr}{m} \]

### Step 6: Calculate the kinetic energy of the alpha particle
The kinetic energy \( E \) of the alpha particle can be expressed as:
\[ E = \frac{1}{2}mv^2 \]
Substituting the expression for \( v \) into the kinetic energy formula:
\[ E = \frac{1}{2}m\left(\frac{qBr}{m}\right)^2 \]
This simplifies to:
\[ E = \frac{1}{2} \frac{q^2B^2r^2}{m} \]

### Step 7: Interpret the result
This equation gives us the kinetic energy of the alpha particle in terms of its charge \( q \), the magnetic field \( B \), the radius of the circular path \( r \), and its mass \( m \).