Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 46, Problem 19P

(a)

To determine

The Q value of the decay of Λ0 particle.

(a)

Expert Solution
Check Mark

Answer to Problem 19P

The Q value of the decay reaction is 37.7MeV.

Explanation of Solution

The Λ0 particle decay into a π meson. Write the decay reaction, keeping in mind the required conservation laws.

    Λ0p+π

Write the formula for the Q value of the above decay reaction.

    Q=mΛc2mPc2mπc2

Here, Q is the Q value of the decay reaction, mΛc2 is the rest energy of Λ0, mPc2 is the rest energy of proton, and mπc2 is the rest mass of the π.

Conclusion:

Substitute 1115.6MeV for mΛc2, 938.3MeV for mPc2, 139.6MeV for mπc2.

    Q=1115.6MeV938.3MeV139.6MeV=37.7MeV

The Q value of the decay reaction is 37.7MeV.

(b)

To determine

The total kinetic energy shared by the proton and the π meson.

(b)

Expert Solution
Check Mark

Answer to Problem 19P

The kinetic energy of the products is 37.7MeV.

Explanation of Solution

The Q value is the difference in the kinetic energy of the reactants and the products. The kinetic energy of the reactant is zero, thus the kinetic energy of the products will be equal to the Q value of the reaction.

From section (a) the Q value of the reaction is 37.7MeV. Thus the kinetic energy of the products is 37.7MeV.

Conclusion:

The kinetic energy of the products is 37.7MeV.

(c)

To determine

The total momentum shared by the proton and the π meson.

(c)

Expert Solution
Check Mark

Answer to Problem 19P

The total momentum shared by the proton and the π meson will be zero.

Explanation of Solution

The Λ0 particle is at rest. So the momentum of the reactant is zero. As per the conservation of momentum, the total momentum of the products have to be zero in order to conserve the momentum.

Thus the total momentum shared by the proton and the π meson will be zero.

Conclusion:

The total momentum shared by the proton and the π meson will be zero.

(d)

To determine

Whether the proton and the π meson have same kinetic energy.

(d)

Expert Solution
Check Mark

Answer to Problem 19P

The π meson has more kinetic energy than the proton.

Explanation of Solution

The momentum of the proton and π meson is same. However, as their masses are different, their kinetic energy will be different. Since the mass of π meson is too less than the mass of proton, the π meson will have more kinetic energy than the proton.

Write the formula for the kinetic energy of the proton.

    Kp=mp2c4+p2c2mpc2                                                                                  (I)

Here, Kp is the kinetic energy of the proton, mp is the mass of the proton, c is the speed of light in vacuum, and p is the momentum.

Write the formula for the kinetic energy of the π meson.

    Kπ=mπ2c4+p2c2mπc2                                                                                    (II)

Here, Kπ is the kinetic energy of the π meson, and mπ is the mass of π meson.

The sum of kinetic energy of the proton and π meson is equal to the Q value of the reaction.

    mp2c4+p2c2mpc2+mπ2c4+p2c2mπc2=Q                                                  =mΛc2mpc2mπc2

Solve the above equation.

    mp2c4+p2c2=mΛc2mπ2c4+p2c2mp2c4+p2c2=mΛ2c42mΛc2mπ2c4+p2c2+mπ2c4+p2c2mπ2c4+p2c2=mΛ2c4mp2c4+mπ2c42mΛc2pc=(mΛ2c4mp2c4+mπ2c42mΛc2)2mπ2c4                                   (III)

Substitute 1115.6MeV for mΛc2, 938.3MeV for mpc2, 139.6MeV for mπc2 in equation (III).

    pc=(1115.62938.32+139.622(1115.6)MeV)2(139.6MeV)2=100.4MeV

Substitute 100.4MeV for pc, 938.3MeV for mpc2 in equation (I).

    Kp=(938.3MeV)2+(100.4MeV)2938.3MeV=5.35MeV

Substitute 100.4MeV for pc, 139.6MeV for mπc2 in equation (II).

    Kπ=(139.6MeV)2+(100.4MeV)2139.6MeV=32.3MeV

Thus the π meson has more kinetic energy than the proton.

Conclusion:

The π meson has more kinetic energy than the proton.

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Chapter 46 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

Ch. 46 - Prob. 6OQCh. 46 - Prob. 7OQCh. 46 - Prob. 8OQCh. 46 - Prob. 1CQCh. 46 - Prob. 2CQCh. 46 - Prob. 3CQCh. 46 - Prob. 4CQCh. 46 - Prob. 5CQCh. 46 - Prob. 6CQCh. 46 - Prob. 7CQCh. 46 - Prob. 8CQCh. 46 - Prob. 9CQCh. 46 - Prob. 10CQCh. 46 - Prob. 11CQCh. 46 - Prob. 12CQCh. 46 - Prob. 13CQCh. 46 - Prob. 1PCh. 46 - Prob. 2PCh. 46 - Prob. 3PCh. 46 - Prob. 4PCh. 46 - Prob. 5PCh. 46 - Prob. 6PCh. 46 - Prob. 7PCh. 46 - Prob. 8PCh. 46 - Prob. 9PCh. 46 - Prob. 10PCh. 46 - Prob. 11PCh. 46 - Prob. 12PCh. 46 - Prob. 13PCh. 46 - Prob. 14PCh. 46 - Prob. 15PCh. 46 - Prob. 16PCh. 46 - Prob. 17PCh. 46 - Prob. 18PCh. 46 - Prob. 19PCh. 46 - Prob. 20PCh. 46 - Prob. 21PCh. 46 - Prob. 22PCh. 46 - Prob. 23PCh. 46 - Prob. 24PCh. 46 - Prob. 25PCh. 46 - Prob. 26PCh. 46 - Prob. 27PCh. 46 - Prob. 28PCh. 46 - Prob. 29PCh. 46 - Prob. 30PCh. 46 - Prob. 31PCh. 46 - Prob. 32PCh. 46 - Prob. 33PCh. 46 - Prob. 34PCh. 46 - Prob. 35PCh. 46 - Prob. 36PCh. 46 - Prob. 37PCh. 46 - Prob. 38PCh. 46 - Prob. 39PCh. 46 - Prob. 40PCh. 46 - Prob. 41PCh. 46 - Prob. 42PCh. 46 - Prob. 43PCh. 46 - Prob. 44PCh. 46 - The various spectral lines observed in the light...Ch. 46 - Prob. 47PCh. 46 - Prob. 48PCh. 46 - Prob. 49PCh. 46 - Prob. 50PCh. 46 - Prob. 51APCh. 46 - Prob. 52APCh. 46 - Prob. 53APCh. 46 - Prob. 54APCh. 46 - Prob. 55APCh. 46 - Prob. 56APCh. 46 - Prob. 57APCh. 46 - Prob. 58APCh. 46 - An unstable particle, initially at rest, decays...Ch. 46 - Prob. 60APCh. 46 - Prob. 61APCh. 46 - Prob. 62APCh. 46 - Prob. 63APCh. 46 - Prob. 64APCh. 46 - Prob. 65APCh. 46 - Prob. 66APCh. 46 - Prob. 67CPCh. 46 - Prob. 68CPCh. 46 - Prob. 69CPCh. 46 - Prob. 70CPCh. 46 - Prob. 71CPCh. 46 - Prob. 72CPCh. 46 - Prob. 73CP
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