Chapter 26, Problem 38A

(a)

To determine

The speed of the particle.

Answer to Problem 38A

  $\boxed{v=8.73\times {10}^5\text{m/s}}$ .

Explanation of Solution

Given:

Mass of an alpha particle is $m=6.6\times {10}^{-27}\text{kg}$ .

Charge of an alpha particle is $q=2e$ .

Strength of magnetic field is $B=0.20\text{T}$ .

Radius of the circular path is $r=0.090\text{m}$ .

Formula:

The relation between magnetic force and centripetal force is

  $Bqv=\frac{m{v}^2}{r}$  ...... (1)

Here, $B,q,v,m$ and $r$ are strength of the magnetic field, charge of the particle, velocity of the particle, mass of an alpha particle and radius of the circular path.

Calculation:

Rearranging the equation (1)

  $v=\frac{Bqr}{m}$  ...... (2)

Substitute $2\times \left(1.6\times {10}^{-19}\text{C}\right)$ for $q$ , $6.6\times {10}^{-27}\text{kg}$ for $m$ , $0.09\text{m}$ for $r$ , $0.2\text{T}$ for $B$ in equation (2)

  $\begin{array}{l}v=\frac{\left(0.20\text{T}\right)\times \left(2\times \left(1.6\times {10}^{-19}\text{C}\right)\right)\times \left(0.09\text{m}\right)}{6.6\times {10}^{-27}\text{kg}}\\ \boxed{v=8.73\times {10}^5\text{m/s}}\end{array}$

Conclusion:

Hence, the speed of alpha particle is $\boxed{v=8.73\times {10}^5\text{m/s}}$ .

(b)

To determine

The kinetic energy of the particle.

Answer to Problem 38A

  $\boxed{\text{KE}=2.51\times {10}^{-15}\text{J}}$ .

Explanation of Solution

Given:

The speed of alpha particle is $v=8.73\times {10}^5\text{m/s}$ .

Mass of an alpha particle is $m=6.6\times {10}^{-27}\text{kg}$ .

Formula used:

The expression for the kinetic energy is

  $\text{KE}=\frac{1}{2}m{v}^2$  ...... (2)

Here, $m$ and $v$ are mass and speed of the particle.

Calculation:

Substitute $6.6\times {10}^{-27}\text{kg}$ for $m$ and $8.73\times {10}^5\text{m/s}$ for $v$ in equation (2)

  $\begin{array}{l}\text{KE}=\frac{1}{2}\times \left(6.6\times {10}^{-27}\text{kg}\right)\times {\left(8.73\times {10}^5\text{m/s}\right)}^2\\ \boxed{\text{KE}=2.51\times {10}^{-15}\text{J}}\end{array}$

Conclusion:

Hence, the required kinetic energy of the particle is $\boxed{\text{KE}=2.51\times {10}^{-15}\text{J}}$ .

(c)

To determine

The potential difference that would be required to give the required speed.

Answer to Problem 38A

  $\boxed{V=7.8\times {10}^3\text{Volt}}$ .

Explanation of Solution

Given:

Kinetic energy of the particle is $\text{KE}=2.51\times {10}^{-15}\text{J}$ .

Charge of an alpha particle is $q=2e$ .

Formula used:

The kinetic energy of a particle that is accelerated from rest through a potential difference can be expressed as

  $\text{KE}=qV$  ...... (1)

Here, $q$ and $V$ are charge and potential difference.

Calculation:

Rearranging equation (1)

  $V=\frac{\text{KE}}{q}$  ...... (2)

Substitute $2.51\times {10}^{-15}\text{J}$ for $\text{KE}$ and $2e$ for $q$ in equation (2)

  $\begin{array}{l}V=\frac{\left(2.51\times {10}^{-15}\text{J}\right)}{\left(2e\right)}\\ V=\frac{\left(2.51\times {10}^{-15}\text{J}\right)}{\left(2\times 1.6\times {10}^{-19}\text{C}\right)}\\ \boxed{V=7.8\times {10}^3\text{Volt}}\end{array}$

Conclusion:

Hence, the required potential difference is $\boxed{V=7.8\times {10}^3\text{Volt}}$ .