© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the relationship between radius and motion in a magnetic field
When a charged particle (mass $m$, charge $q$, velocity $v$) moves in a uniform magnetic field $B$, it travels in a circular path. The centripetal force required for circular motion is provided by the magnetic force:
$F_{\text{centripetal}} = F_{\text{magnetic}} \quad \Rightarrow \quad \frac{m v^2}{R} = q\, v\, B.$
Cancelling $v$ from both sides (assuming $v \neq 0$):
$\displaystyle R = \frac{m v}{q B}.$
Step 2: Express the radius in terms of kinetic energy
The kinetic energy $K$ of a particle is given by:
$\displaystyle K = \frac{1}{2} m v^2 \quad \Longrightarrow \quad v = \sqrt{\frac{2K}{m}}.$
Substituting $v$ in the expression for $R$:
$\displaystyle R = \frac{m}{qB} \sqrt{\frac{2K}{m}} = \frac{\sqrt{2mK}}{qB}.$
Step 3: Apply the formula to a proton
For a proton (mass $m$, charge $q$) with kinetic energy $K_p = 1\ \text{MeV}$, the radius of the circular path is:
$\displaystyle R_p = \frac{\sqrt{2\,m\,K_p}}{q\,B}.$
Step 4: Apply the formula to an α-particle
An α-particle (which is a helium nucleus) has:
Mass $= 4m$ (approximately 4 times the proton mass, if $m$ is the proton mass),
Charge $= 2q$ (twice the charge of the proton).
Let the kinetic energy of the α-particle be $K_\alpha$. Then its velocity is:
$\displaystyle v_\alpha = \sqrt{\frac{2\,K_\alpha}{4m}} = \sqrt{\frac{K_\alpha}{2m}},$
and so the radius of its path is:
$\displaystyle R_\alpha = \frac{\sqrt{2\,(4m)\,K_\alpha}}{(2q)\,B} = \frac{\sqrt{8\,m\,K_\alpha}}{2\,q\,B} = \frac{\sqrt{2\,m\,K_\alpha}}{q\,B}.$
Step 5: Equate the radii for the same magnetic field
For the α-particle to move in the same circle of radius $R$ as the proton, we must have $R_p = R_\alpha$. From the expressions:
$\displaystyle R_p = \frac{\sqrt{2\,m\,K_p}}{q\,B}, \quad R_\alpha = \frac{\sqrt{2\,m\,K_\alpha}}{q\,B}.$
Setting $R_p = R_\alpha$:
$\displaystyle \sqrt{2\,m\,K_p} = \sqrt{2\,m\,K_\alpha} \quad \Longrightarrow \quad K_p = K_\alpha.$
Step 6: Substitute the known kinetic energy of the proton
We know the proton’s kinetic energy $K_p$ is $1 \ \text{MeV}$. Therefore, $K_\alpha$ must also equal $1 \ \text{MeV}$ to maintain the same radius in the same magnetic field.
Final Answer
The energy of the α-particle should be $1 \ \text{MeV}$.
