Answer any two numerical questions. a) An ion for which the charge per unit mass is 4.40 10^7\,C/kg has velocity of 3.52 10^5\,m/s and moves in a circular orbit in a magnetic field of flux density…

a) Radius of the Circular Orbit in a Magnetic Field

Given:

• Charge per unit mass of the ion, $\frac{q}{m}=4.40\times 10^{7}C/kg$
• Velocity of the ion, $v=3.52\times 10^{5}m/s$
• Magnetic flux density, $B=0.4\mathrm{T}$

Objective: Find the radius $r$ of the circular orbit.

Solution: When a charged particle moves in a magnetic field, it experiences a centripetal force due to the magnetic Lorentz force. The magnetic force acts as the centripetal force, and the radius of the orbit can be determined using the following relation:

$$\frac{m{v}^2}{r}=qvB$$

Rearranging for $r$:

$$r=\frac{mv}{qB}$$

Since $\frac{q}{m}$ is given, we can rewrite the equation as:

$$r=\frac{v}{\left(\frac{q}{m}\right)B}$$

Substitute the given values:

$$r=\frac{3.52\times 10^{5}m/s}{(4.40\times 10^{7}C/kg)\times (0.4\mathrm{T})}$$

Calculate the denominator:

$$(4.40\times 10^7)\times 0.4=1.76\times 10^{7}C/kg\cdot \mathrm{T}$$

Now, divide the velocity by the denominator:

$$r=\frac{3.52\times 10^5}{1.76\times 10^7}=0.02\mathrm{m}$$

Final Answer: The radius of the orbit is 0.02 m.

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b) De Broglie Wavelength of a Neutron

Given:

• Kinetic energy of the neutron, $K=150\mathrm{eV}$
• Mass of the neutron, $m=1.675\times 10^{-27}\mathrm{kg}$
• Planck's constant, $h=6.6\times 10^{-34}\mathrm{Js}$
• Conversion factor, $1\mathrm{eV}=1.6\times 10^{-19}\mathrm{J}$

Objective: Find the de Broglie wavelength $\lambda$ of the neutron.

Solution: The de Broglie wavelength is given by:

$$\lambda =\frac{h}{p}$$

where $p$ is the momentum of the neutron. The momentum can be found using the kinetic energy:

$$K=\frac{p^2}{2m}$$

Rearranging for $p$:

$$p=\sqrt{2mK}$$

First, convert the kinetic energy from eV to Joules:

$$K=150\times 1.6\times 10^{-19}=2.4\times 10^{-17}\mathrm{J}$$

Now, calculate the momentum:

$$p=\sqrt{2\times 1.675\times 10^{-27}\times 2.4\times 10^{-17}}$$

$$p=\sqrt{8.064\times 10^{-44}}=8.98\times 10^{-22}\mathrm{kg}\cdot m/s$$

Now, calculate the de Broglie wavelength:

$$\lambda =\frac{6.6\times 10^{-34}}{8.98\times 10^{-22}}=7.35\times 10^{-13}\mathrm{m}$$

Final Answer: The de Broglie wavelength of the neutron is $7.35\times 10^{-13}$ m.

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c) Binding Energy per Nucleon of ${}_{}^{}{}_{26}^{56}\mathrm{Fe}$

Given:

• Atomic mass of ${}_{}^{}{}_{26}^{56}\mathrm{Fe}$, $M_{\text{Fe}}=55.9349\mathrm{u}$
• Atomic mass of ${}_{}^{}{}_1^1\mathrm{H}$, $M_{\text{H}}=1.00783\mathrm{u}$
• Mass of neutron, $M_n=1.00867\mathrm{u}$
• Conversion factor, $1\mathrm{u}=931\mathrm{MeV}$

Objective: Calculate the binding energy per nucleon of ${}_{}^{}{}_{26}^{56}\mathrm{Fe}$.

Solution: The binding energy per nucleon is calculated using the mass defect formula:

$$\text{Mass defect}=\left[Z\cdot M_{\text{H}}+(A-Z)\cdot M_n\right]-M_{\text{Fe}}$$

where:

• $Z=26$ (number of protons)
• $A=56$ (mass number)

Calculate the total mass of protons and neutrons separately:

$$\text{Total mass of protons}=26\times 1.00783=26.20358\mathrm{u}$$

$$\text{Total mass of neutrons}=(56-26)\times 1.00867=30\times 1.00867=30.2601\mathrm{u}$$

Now, calculate the total mass of protons and neutrons:

$$\text{Total mass}=26.20358+30.2601=56.46368\mathrm{u}$$

Calculate the mass defect:

$$\text{Mass defect}=56.46368-55.9349=0.52878\mathrm{u}$$

Convert the mass defect to energy using $1\mathrm{u}=931\mathrm{MeV}$:

$$\text{Binding energy}=0.52878\times 931=491.58\mathrm{MeV}$$

Calculate the binding energy per nucleon:

$$\text{Binding energy per nucleon}=\frac{491.58}{56}=8.78MeV/nucleon$$

Final Answer: The binding energy per nucleon of ${}_{}^{}{}_{26}^{56}\mathrm{Fe}$ is 8.78 MeV/nucleon.