Chapter 32, Problem 6P

To determine

(a)

What maximum kinetic energy ($K$)and speed $(v)$ could alpha particle attain?

Answer to Problem 6P

Solution:

The maximum kinetic energy of alpha particle is 8.6 MeV and reaches a speed of 2.0x10⁷ m/s.

Explanation of Solution

Given Info:

The charge $\left(q_{\alpha }\right)$ of alpha particle is twice the charge of proton . The mass $\left(m_{\alpha }\right)$ of alpha particle is four times proton’s mass. The Cyclotron’s maximum radius $\left(R\right)$ is 0.25 m and the magnetic field $\left(B\right)$ is 1.7 T.

Formula used:

The maximum kinetic energy $\left(K_{max}\right)$ can be defined as:

$K_{max}=\frac{1}{2}{m}_{\alpha }{v}_{max}^2$ (1)

Where:

• $m$ alpha particle’s mass, which is $m_{alpha}=4m$, with $m$ the proton’s mass

• $v_{max}$ maximum speed reached

Similarly, the max speed $\left(v_{max}\right)$ in a circular motion caused by a magnetic field $\left(B\right)$ can be obtained using:

$v_{max}=\frac{q_{\alpha }.B.R_{max}}{m_{\alpha }}$ (2)

Where:

• $q_{\alpha }$ alpha particle’s charge, which is: $q_{\alpha }=2q$, with $q$ the proton’s charge

• $B$ magnetic field applied

• $R_{max}$ maximum radius of the cyclotron

Calculation:

Replacing (2)into (1)we obtain:

$K_{max}=\frac{q_{\alpha }{}^2{B}^2{R}_{max}^2}{2m_{\alpha }}=\frac{4q^2{B}^2{R}_{max}^2}{8m}=\frac{q^2{B}^2{R}_{max}^2}{2m}$ (3)

Replacing data given values into equation (3)we get:

$K_{max}=\frac{{(1.6\times {10}^{-19}C)}^2{(1.7T)}^2{(0.25m)}^2}{2\times 1.67\times {10}^{-27}kg}$ (4)

Solving we obtain:

$K_{max}=1.38\times {10}^{-12}J$ (5)

Converting to MeV we get:

$K_{max}=1.38\times {10}^{-12}J.\left(\frac{6.24\times {10}^{12}MeV}{1J}\right)=8.6MeV$ (6)

Now, the value of speed can be obtained isolating $v_{max}$ from equation (1):

$v_{max}=\sqrt{\frac{2K_{max}}{m_{\alpha }}}=\sqrt{\frac{2K_{max}}{4m}}=\sqrt{\frac{K_{max}}{2m}}$ (7)

Replacing (5)and given data values we obtain:

$v_{max}=\sqrt{\frac{(1.38\times {10}^{-12}J)}{2(1.67\times {10}^{-27}kg)}}=2.0\times {10}^{7}m/s$ (8)

To determine

(b)

Same as part (a)but use deuterons $\left({}_1{}^{2}H\right)$

Answer to Problem 6P

Solution:

Explanation of Solution

Given info:

Mass of deuterons $\left(m_d\right)$ is $2m_{proton}$ and charge of deuterons $\left(q_d\right)$ is equal to the charge of proton.The Cyclotron’s maximum radius $\left(R\right)$ is 0.25 m and the magnetic field $\left(B\right)$ is 1.7 T.

Formula used:

According to equation (3)we have:

$K_{max}=\frac{q_d^2{B}^2{R}_{max}^2}{2m_d}=\frac{q^2{B}^2{R}_{max}^2}{4m}$ (9)

Similarly, from equation (7), the speed can be obtained using:

$v_{max}=\sqrt{\frac{2K_{max}}{m_d}}=\sqrt{\frac{2K_{max}}{2m}}=\sqrt{\frac{K_{max}}{m}}$ (10)

Calculation:

Inserting given data into equation (9)we have:

$K_{max}=\frac{{(1.60\times {10}^{-19}C)}^2{(1.7T)}^2{(0.25m)}^2}{4(1.67\times {10}^{-27}kg)}$ (11)

Solving we obtain:

$K_{max}=6.92\times {10}^{-13}J$ (12)

Converting to MeV we get:

$K_{max}=6.92\times {10}^{-13}J.\left(\frac{6.24\times {10}^{12}MeV}{1J}\right)=4.3MeV$ (13)

Similarly, for speedreplace given data into equation (10):

$v_{max}=\sqrt{\frac{6.92\times {10}^{-13}J}{1.67\times {10}^{-27}kg}}=2.0\times {10}^{7}m/s$ (14)

Conclusion:

According to results (13)and (14)the maximum kinetic energy of deuterons is 4.3 MeV and reaches a speed of 2.0x10⁷ m/s.

To determine

(c)

In each case,what frequency ($f$)of voltage is required.

Answer to Problem 6P

Solution:

The frequency of voltage required is 13 MHz for both cases.

Explanation of Solution

Given info:

The magnetic field $\left(B\right)$ is 1.7 T. The charge $q_{\alpha }$ and mass $m_{\alpha }$ of alpha particle are respectively: $2q$ and $4m$. The charge $q_d$ and mass $m_d$ of deuterons are respectively: $q$ and $2m$.

Formula used:

The cyclotron frequency($f$)is defined as:

$f=\frac{qB}{2\pi m}$ (15)

Calculation:

For alpha particle we have:

$f_{\alpha }=\frac{q_{\alpha }B}{2\pi m_{\alpha }}$ (16)

But:

$q_{\alpha }=2q$ (17)

And

$m_{\alpha }=4m$ (18)

Inserting (17)and (18)we get:

$f_{\alpha }=\frac{qB}{4\pi m}$ (19)

Replacing given data values we obtain:

$f_{\alpha }=\frac{(1.6\times {10}^{-19})(1.7T)}{4\pi (1.67\times {10}^{-27}kg)}=1.3\times {10}^{7}Hz=13MHz$ (20)

Similarly, for deuterium we have:

$f_d=\frac{q_{d}B}{2\pi m_d}$ (21)

$q_d=q$ (22)

And

$m_d=2m$ (23)

Replacing (22)and (23)into (21)we get:

$f_d=\frac{qB}{4\pi m}$ (24)

Which is the same as equation (19), therefore; the cyclotron frequency for deuterium is:

$f_d=13MHz$ (25)