Questions & Answers

Question

Answers

A. 9 MeV

B. 10 MeV

C. 7 MeV

D. 11 MeV

Answer
Verified

Hint: We will find out the mass of the proton from the given frequency and magnetic field. Then we will find an expression for the kinetic energy of particles coming out of the oscillator. Finally converting the energy into eV, we can have our answer.

Formula used:

$f=\dfrac{qB}{2\pi m}$

$E_k=\dfrac{q^2B^2R^2}{2m}$

Complete step by step solution:

The frequency of oscillation in a cyclotron is given by,

$f=\dfrac{qB}{2\pi m}$

Now, we know the frequency and the magnetic field B. From question, $B=0.66 T$ and $f=10 MHz$. Again, charge of a proton is equal and opposite to the charge of an electron and is given as $q=1.6 \times 10^{-19} C$ .

So, after putting all this values in the above formula, we obtain the mass of the proton as,

$m=\dfrac{qB}{2\pi f}=1.68\times 10^{-27}kg$

Now, the condition for uninterrupted oscillation in the cyclotron is that the magnetic force on the proton has to be equal to the centrifugal force of its rotation. So,

$qvB=\dfrac{mv^2}{r}$

Here, v is the velocity of the proton and r is the radius of its orbit. After a little modification the formula becomes, $mv=qBr$. It gives the momentum of the proton.

Now, when the proton is about to leave the dee, $r=R$. Here R is the radius of the dee and it’s about 60 cm or 0.6 m here. So, the final kinetic energy is given by,

$E_k=\dfrac{(mv)^2}{2m}=\dfrac{q^2B^2R^2}{2m}=1.19\times 10^{-12}J=7.47 MeV$

Hence, among the given options, option C is the closest to the actual value. Hence, option C is the correct answer.

Additional information:

We cannot accelerate electrons in the normal cyclotron. An electron in high velocity changes its mass and this destroys the equilibrium condition. So, a fixed frequency cyclotron cannot accelerate electrons.

Note: Remember few things,

1. Make sure you convert energy from Joule to electron volt by dividing it by electronic charge.

2. In a formula, keep all the quantities in the same unit system.

3. If you remember the mass of the proton, you don’t need to calculate it again.

Formula used:

$f=\dfrac{qB}{2\pi m}$

$E_k=\dfrac{q^2B^2R^2}{2m}$

Complete step by step solution:

The frequency of oscillation in a cyclotron is given by,

$f=\dfrac{qB}{2\pi m}$

Now, we know the frequency and the magnetic field B. From question, $B=0.66 T$ and $f=10 MHz$. Again, charge of a proton is equal and opposite to the charge of an electron and is given as $q=1.6 \times 10^{-19} C$ .

So, after putting all this values in the above formula, we obtain the mass of the proton as,

$m=\dfrac{qB}{2\pi f}=1.68\times 10^{-27}kg$

Now, the condition for uninterrupted oscillation in the cyclotron is that the magnetic force on the proton has to be equal to the centrifugal force of its rotation. So,

$qvB=\dfrac{mv^2}{r}$

Here, v is the velocity of the proton and r is the radius of its orbit. After a little modification the formula becomes, $mv=qBr$. It gives the momentum of the proton.

Now, when the proton is about to leave the dee, $r=R$. Here R is the radius of the dee and it’s about 60 cm or 0.6 m here. So, the final kinetic energy is given by,

$E_k=\dfrac{(mv)^2}{2m}=\dfrac{q^2B^2R^2}{2m}=1.19\times 10^{-12}J=7.47 MeV$

Hence, among the given options, option C is the closest to the actual value. Hence, option C is the correct answer.

Additional information:

We cannot accelerate electrons in the normal cyclotron. An electron in high velocity changes its mass and this destroys the equilibrium condition. So, a fixed frequency cyclotron cannot accelerate electrons.

Note: Remember few things,

1. Make sure you convert energy from Joule to electron volt by dividing it by electronic charge.

2. In a formula, keep all the quantities in the same unit system.

3. If you remember the mass of the proton, you don’t need to calculate it again.

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