
A proton of a mass m and charge \[ + e\] is moving in a circular orbit of a magnetic field with energy \[1{\rm{ }}MeV\]. What should be the energy of \[\alpha \] particle (mass \[4m\] and charge \[ + 2e\]) so that it can revolve in the path of the same radius?
A. \[1{\rm{ }}MeV\]
B. \[4{\rm{ }}MeV\]
C. \[2{\rm{ }}MeV\]
D. \[0.5{\rm{ }}MeV\]
Answer
161.1k+ views
Hint: In the given question, we need to find the energy of the \[\alpha \] particle so that it revolves in the path of the same radius. For this, we have to use the following formula of radius of circular orbit. That means, we have to find the ratio of kinetic energy of a proton to kinetic energy of \[\alpha \] particles.
Formula used:
The following formula is used for solving the given question.
The radius of the circular path is given by
\[r = \dfrac{{\sqrt {2mK} }}{{qB}}\]
Here, \[r\] is radius of circular orbit, \[m\] is mass of particles, \[K\] kinetic energy, \[q\] is charge and \[B\] is magnetic field strength.
Complete answer:
We know that The radius of the circular path is \[r = \dfrac{{\sqrt {2mK} }}{{qB}}\]
This indicates that \[K\alpha \dfrac{{{q^2}}}{m}\]
So, kinetic energy of a particle is directly proportional to the square of charge and inversely proportional to the mass of a particle.
Now, we will take the ratio of kinetic energy of a proton to the kinetic energy of \[\alpha \] particles.
Mathematically, it is given by
That means, \[\dfrac{{{K_p}}}{{{K_\alpha }}} = \dfrac{{{q_p}^2}}{{{q_\alpha }^2}} \times \dfrac{{{m_\alpha}}}{{{m_p }}}\]
This gives, \[\dfrac{1}{{{K_\alpha }}} = {\left( {\dfrac{e}{{2e}}} \right)^2} \times \dfrac{{4m}}{m}\]
By simplifying, we get
\[\dfrac{1}{{{K_\alpha }}} = 1\]
So, we get
\[{K_\alpha } = 1{\rm{ }}MeV\]
Hence, the energy of \[\alpha \]particle is \[1{\rm{ }}MeV\].
Therefore, the correct option is (A).
Note: Many students make mistakes in the calculation part as well as writing the formula of between radius of circular path for a particle. This is the only way through which we can solve the example in the simplest way. Also, it is essential to analyze the result of relation to get the desired result.
Formula used:
The following formula is used for solving the given question.
The radius of the circular path is given by
\[r = \dfrac{{\sqrt {2mK} }}{{qB}}\]
Here, \[r\] is radius of circular orbit, \[m\] is mass of particles, \[K\] kinetic energy, \[q\] is charge and \[B\] is magnetic field strength.
Complete answer:
We know that The radius of the circular path is \[r = \dfrac{{\sqrt {2mK} }}{{qB}}\]
This indicates that \[K\alpha \dfrac{{{q^2}}}{m}\]
So, kinetic energy of a particle is directly proportional to the square of charge and inversely proportional to the mass of a particle.
Now, we will take the ratio of kinetic energy of a proton to the kinetic energy of \[\alpha \] particles.
Mathematically, it is given by
That means, \[\dfrac{{{K_p}}}{{{K_\alpha }}} = \dfrac{{{q_p}^2}}{{{q_\alpha }^2}} \times \dfrac{{{m_\alpha}}}{{{m_p }}}\]
This gives, \[\dfrac{1}{{{K_\alpha }}} = {\left( {\dfrac{e}{{2e}}} \right)^2} \times \dfrac{{4m}}{m}\]
By simplifying, we get
\[\dfrac{1}{{{K_\alpha }}} = 1\]
So, we get
\[{K_\alpha } = 1{\rm{ }}MeV\]
Hence, the energy of \[\alpha \]particle is \[1{\rm{ }}MeV\].
Therefore, the correct option is (A).
Note: Many students make mistakes in the calculation part as well as writing the formula of between radius of circular path for a particle. This is the only way through which we can solve the example in the simplest way. Also, it is essential to analyze the result of relation to get the desired result.
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