Chapter 29, Problem 93P

(a)

To determine

The speed of the $\alpha$ particle with kinetic energy of $4.17\text{ MeV}$.

Answer to Problem 93P

The speed of the $\alpha$ particle with kinetic energy of $4.17\text{ MeV}$ is $1.40\times {10}^7\text{m}/\text{s}$.

Explanation of Solution

${}^{238}\text{U}$ undergoes radioactive decay and produces an $\alpha$ particle with kinetic energy of $K=4.17\text{ MeV}$. The atomic mass of an $\alpha$ particle is approximately $4.00\text{ u}$.

Write the formula for kinetic energy

    $K=\frac{1}{2}m{v}^2$ (I)

Here, $K$ is the kinetic energy, $m$ is the mass of the particle and $v$ is the velocity of the particle.

Conclusion:

Substitute $4.00\text{ u}$ for $m$ and $4.17\text{ MeV}$ for $K$ in equation (I) to find the value of $v$. [Note: $1\text{ u}=1.66\times {10}^{-27}\text{ kg}$ and $1\text{ eV}=1.602\times {10}^{-19}\text{ J}$]

$\begin{array}{l}v=\sqrt{2K/m}\\ v=\sqrt{2\times 4.17\times {10}^6\text{ eV}/4.00\text{ u}}\\ v=\sqrt{\frac{2\times 4.17\times {10}^6\text{ eV}\times 1.602\times {10}^{-19}\text{J}/\text{eV}}{4.00\text{ u}\times 1.66\times {10}^{-27}\text{kg}/\text{u}}}\\ v=1.40\times {10}^7\text{m}/\text{s}\end{array}$

Thus, the speed of the $\alpha$ particle with kinetic energy of $4.17\text{ MeV}$ is $1.40\times {10}^7\text{m}/\text{s}$.

(b)

To determine

The electric field that would allow the $\alpha$ particle to pass through the velocity selector undeflected.

Answer to Problem 93P

The electric field that would allow the $\alpha$ particle to pass through the velocity selector undeflected is $4.3\times {10}^6\text{V}/\text{m}$.

Explanation of Solution

The magnet produces a magnetic field of $B=0.30\text{ T}$ and the speed of the $\alpha$ particle is $v=1.40\times {10}^7\text{m}/\text{s}$. To pass through the velocity selector undeflected the electric and magnetic force should be equal, so the net force on the $\alpha$ particle is zero.

Write the formula for magnetic force

    $F_M=qvB$ (II)

Here, $F_M$ is the magnetic force, $q$ is the charge, $v$ is the velocity and $B$ is the magnetic field.

Write the formula for electric force

    $F_E=qE$                                                                                            (III)

Here, $F_E$ is the electric force, $E$ is the electric field.

Conclusion:

Equating equation (I) and (II) to solve for $E$

$\begin{array}{c}{F}_M=qvB=F_E=qE\\ qvB=qE\\ E=Bv\end{array}$

Substitute $1.40\times {10}^7\text{m}/\text{s}$ for $v$ and $0.30\text{ T}$ for $B$ in the above equation to solve for $E$

$\begin{array}{l}E=0.30\text{ T}\times 1.42\times {10}^7\text{m}/\text{s}\\ E=4.3\times {10}^6\text{V}/\text{m}\end{array}$

Thus, The electric field that would allow the $\alpha$ particle to pass through the velocity selector undeflected is $4.3\times {10}^6\text{V}/\text{m}$.

(c)

To determine

The radius of the trajectory of an $\alpha$ particle in a $0.30\text{T}$ field.

Answer to Problem 93P

The radius of the trajectory of an $\alpha$ particle in a $0.30\text{T}$ field is $98\text{ cm}$.

Explanation of Solution

The magnet produces a magnetic field of $B=0.30\text{ T}$ and the speed of the $\alpha$ particle is $v=1.40\times {10}^7\text{m}/\text{s}$. The atomic mass of an $\alpha$ particle is approximately $4.00\text{ u}$.

Write the formula for force [Newton’s second law]

    $F=ma$ (IV)

Here, $F$ is the force and $a$ is the acceleration of the particle.

Conclusion:

Equating equation (II) and (IV) to solve for r

$\begin{array}{l}F=qvB=ma\\ qvB=m\frac{v^2}{r}\\ r=\frac{mv}{Bq}\end{array}$

Substitute $1.40\times {10}^7\text{m}/\text{s}$ for $v$, $4.00\text{ u}$ for $m$ and $0.30\text{ T}$ for $B$ in the above equation to solve for $r$

$\begin{array}{l}r=\frac{4.00\text{ u}\times 1.40\times {10}^7\text{m}/\text{s}}{0.30\text{ T}}\\ r=\frac{4.00\text{ u}\times 1.66\times {10}^{-27}\text{kg}/\text{u}\times 1.40\times {10}^7\text{m}/\text{s}}{0.30\text{ T}\times \text{1}\text{.602}\times {\text{10}}^{-19}\text{ C}}\\ r=98\text{ cm}\end{array}$

Thus, the radius of the trajectory of an $\alpha$ particle in a $0.30\text{T}$ field is $98\text{ cm}$.

(d)

To determine

Why only the charge-to mass ratio of a particle can be determined using velocity selector, but not the individual values of $q$ and $m$.

Answer to Problem 93P

As both the mass and the charge of the particle affects the radius of trajectory and the values of $v,B\text{ and }r$ can only be measured in this experiment. Hence, only the charge-to mass ratio of a particle can be determined using velocity selector, but not the individual values of $q$ and $m$.

Explanation of Solution

Equating equation (II) and (IV) to find the charge-to-mass ratio

$\begin{array}{l}F=qvB=ma\\ qvB=m\frac{v^2}{r}\\ \frac{q}{m}=\frac{v}{Br}\end{array}$

Here, as both the mass and the charge of the particle affects the radius of trajectory and the values of $v,B\text{ and }r$ can only be measured in this experiment. Hence, only the charge-to mass ratio of a particle can be determined using velocity selector, but not the individual values of $q$ and $m$.