A particle leaving a cyclotron has a total relativistic energy of 10GeVand a relativistic momentum of 8GeV. What is the rest mass of this particle?a)1.20GeV/c2b)0.25GeV/c2c)6GeV/c2d)2.00GeV/c2Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Given:
Total relativistic energy (E) = 10 GeV
Relativistic momentum (p) = 8 GeV

To Find:
Rest mass of the particle

Formula:
The relativistic energy of a particle is given by the equation:
E = √(p^2c^2 + m^2c^4)
where,
E = relativistic energy
p = relativistic momentum
m = rest mass of the particle
c = speed of light in vacuum (3 x 10^8 m/s)

Let's solve the equation to find the rest mass.

Solution:

We are given the relativistic energy (E) and relativistic momentum (p) of the particle. We can use the formula mentioned above to find the rest mass (m).

E = √(p^2c^2 + m^2c^4)

Substituting the given values:

10 GeV = √((8 GeV)^2c^2 + m^2c^4)

Simplifying the equation:

100 GeV^2 = (64 GeV^2c^2 + m^2c^4)

Rearranging the equation:

m^2c^4 = 100 GeV^2 - 64 GeV^2c^2

m^2c^4 = 100 GeV^2 - 64 GeV^2(3 x 10^8 m/s)^2

m^2c^4 = 100 GeV^2 - 64 GeV^2(9 x 10^16 m^2/s^2)

m^2c^4 = 100 GeV^2 - 576 GeV^2 x 10^16 m^2/s^2

m^2c^4 = (100 - 576 x 10^16) GeV^2 x m^2/s^2

m^2c^4 = (-576 x 10^16) GeV^2 x m^2/s^2

Taking the square root on both sides:

mc^2 = √((-576 x 10^16) GeV^2 x m^2/s^2)

mc^2 = √((-576 x 10^16) GeV^2) x √(m^2/s^2)

mc^2 = √((-576 x 10^16) GeV^2) x m/s

mc^2 = √(-576 x 10^16) GeV x m/s

mc^2 = (24i√10^16) GeV x m/s

As we can see, the square root of a negative number is imaginary (i). This implies that the rest mass of the particle is imaginary, which is not physically possible.

Hence, there seems to be an error in the given information or calculations, as the rest mass cannot be determined with the given values.

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