Chapter 36: Q. 42 (page 1060)

One of the important ways in which the Higgs boson was detected at the Large Hadron Collider was by observing a type of decay in which the Higgs—which decays too quickly to be observed directly—is immediately transformed into two photons emitted back to back. Two photons, with momenta $3.31 \times 10^{- 17} k g m / s$ , were detected. What is the mass of the Higgs boson? Give your answer as a multiple of the proton mass.

Short Answer

Expert verified

The mass of the Higgs boson is $m_{H} = 132 m_{p}$

Step by step solution

Given Information

We have given two photons, with momenta $3.31 \times 10^{- 17} k g m / s$

We need to find the mass of the Higgs boson.

Simplify

Consider:

$Thesppedoflight, c = 3 \times 10^{8} m / s Themassofproton, m_{p} = 1.67 \times 10^{- 27} k g Momentumofphoton, p = (3.31 \times 10^{- 17} k g \times m / s)$

Using formula:

$E^{2} - {(p c)}^{2} = {E^{2}}_{0} . . . (1)$

Calculating $E$ from equation (1)

${E^{2}}_{0} = E^{2} - {(p c)}^{2} (W h e r e i s E_{0} = 0 .) \Rightarrow E = 2 p c E = 2 (3.31 \times 10^{- 17} k g \times m / s) \times (3.0 \times 10^{8} m / s) E = 1.99 \times 10^{- 8} J$

Now, finding the Higgs boson :

localid="1652250350638" $E = m_{H} c^{2} m_{H} = \frac{E}{c^{2}} m_{H} = \frac{1.99 \times 10^{- 8} J}{{(3.00 \times 10^{8} m / s)}^{2}} m_{H} = 2.21 \times 10^{- 25} k g$ (Substitute value in the equation )

Mass of the Higgs boson is given as :

localid="1652250363146" $m_{H} = \frac{m H}{m_{p}} m_{H} = \frac{2.21 \times 10^{- 25} k g}{1.67 \times 10^{- 27} k g} m_{H} = 132 m_{p}$

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