Answer any two questions. a) An electron moves in a circular path of radius 20\ cm in a uniform magnetic field of 2 10^ 3T . Find the speed of the electron and period of revolution. Mass of electron…

a) Speed and Period of Revolution of an Electron in a Magnetic Field

Given:

• Radius of circular path, $r=20\mathrm{cm}=0.2\mathrm{m}$
• Magnetic field, $B=2\times 10^{-3}\mathrm{T}$
• Mass of electron, $m=9.1\times 10^{-31}\mathrm{kg}$
• Charge of electron, $q=1.6\times 10^{-19}\mathrm{C}$ (standard value)

Step 1: Find the speed ($v$) of the electron

When an electron moves in a circular path in a magnetic field, it experiences a centripetal force due to the magnetic Lorentz force. The balance of forces gives: $$qvB=\frac{m{v}^2}{r}$$ Solving for $v$: $$v=\frac{qBr}{m}$$ Substitute the given values: $$v=\frac{(1.6\times 10^{-19})(2\times 10^{-3})(0.2)}{9.1\times 10^{-31}}$$ $$v=\frac{6.4\times 10^{-23}}{9.1\times 10^{-31}}=7.03\times 10^{6}m/s$$

Step 2: Find the period ($T$) of revolution

The period is the time taken to complete one revolution. It is related to the speed and radius by: $$T=\frac{2\pi r}{v}$$ Substitute the values: $$T=\frac{2\pi (0.2)}{7.03\times 10^6}=\frac{1.256}{7.03\times 10^6}=1.79\times 10^{-7}\mathrm{s}$$

Final Results:

• Speed of the electron = $7.03\times 10^{6}m/s$
• Period of revolution = $1.79\times 10^{-7}\mathrm{s}$

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b) de Broglie Wavelength of an Electron Accelerated Through a Potential Difference

Given:

• Potential difference, $V=200\mathrm{V}$
• Mass of electron, $m=9.1\times 10^{-31}\mathrm{kg}$
• Planck’s constant, $h=6.6\times 10^{-34}\mathrm{Js}$
• Charge of electron, $q=1.6\times 10^{-19}\mathrm{C}$

Step 1: Find the kinetic energy ($KE$) of the electron

The electron gains kinetic energy equal to the work done by the electric field: $$KE=qV$$ $$KE=(1.6\times 10^{-19})(200)=3.2\times 10^{-17}\mathrm{J}$$

Step 2: Find the momentum ($p$) of the electron

Using the non-relativistic kinetic energy formula: $$KE=\frac{p^2}{2m}$$ Solving for $p$: $$p=\sqrt{2mKE}$$ Substitute the values: $$p=\sqrt{2(9.1\times 10^{-31})(3.2\times 10^{-17})}$$ $$p=\sqrt{5.824\times 10^{-47}}=2.41\times 10^{-23}\mathrm{kg}\cdot m/s$$

Step 3: Find the de Broglie wavelength ($\lambda$)

The de Broglie wavelength is given by: $$\lambda =\frac{h}{p}$$ Substitute the values: $$\lambda =\frac{6.6\times 10^{-34}}{2.41\times 10^{-23}}=2.74\times 10^{-11}\mathrm{m}$$

Final Result:

• de Broglie wavelength = $2.74\times 10^{-11}\mathrm{m}$ or $0.0274\mathrm{nm}$

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c) Activity of 1 g of Radium-226 ($\mathrm{Ra}\text{-}226$)

Given:

• Half-life of $\mathrm{Ra}\text{-}226$, $t_{1/2}=1620\mathrm{years}$
• Mass of sample, $m=1\mathrm{g}$
• Molar mass of $\mathrm{Ra}\text{-}226$, $M=226g/mol$
• Avogadro’s number, $N_A=6.023\times 10^{23}atoms/mol$

Step 1: Find the number of atoms in 1 g of $\mathrm{Ra}\text{-}226$

$$\text{Number of moles}=\frac{m}{M}=\frac{1}{226}$$ $$\text{Number of atoms}=\left(\frac{1}{226}\right)\times 6.023\times 10^{23}=2.665\times 10^{21}\mathrm{atoms}$$

Step 2: Find the decay constant ($\lambda$)

The decay constant is related to the half-life by: $$\lambda =\frac{\ln 2}{t_{1/2}}$$ Convert half-life to seconds (since activity is in Bq, which is decays per second): $$t_{1/2}=1620\times 365\times 24\times 3600=5.11\times 10^{10}\mathrm{s}$$ $$\lambda =\frac{\ln 2}{5.11\times 10^{10}}=1.35\times 10^{-11}{\mathrm{s}}^{-1}$$

Step 3: Find the activity ($A$)

Activity is given by: $$A=\lambda N$$ Substitute the values: $$A=(1.35\times 10^{-11})(2.665\times 10^{21})=3.60\times 10^{10}\mathrm{Bq}$$

Final Result:

• Activity of 1 g of $\mathrm{Ra}\text{-}226$ = $3.60\times 10^{10}\mathrm{Bq}$