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Chapter 19: Problem 166

The speed of daughter nuclei is (A) \(c \sqrt{\frac{\Delta m}{M+\Delta m}}\) (B) \(c \frac{\Delta m}{M+\Delta m}\) (C) \(c \sqrt{\frac{2 \Delta m}{M}}\) (D) \(c \sqrt{\frac{\Delta m}{M}}\)

Short Answer

Expert verified
The derived formula for the speed of daughter nuclei is \(v = c\sqrt{\frac{2螖m}{M - 螖m}}\), which does not match any of the given options. Consult your teacher for clarification.

Step by step solution

01

Apply Conservation of Linear Momentum#

Invariant linear momentum is a fundamental principle in physics. When a nucleus undergoes decay and produces daughter nuclei, the initial and final momentum must be conserved. We can write the momentum conservation equation as: Initial momentum = Final momentum Since the parent nucleus is initially at rest, its initial momentum is zero. After decay, the parent nucleus gets reduced by an amount 螖m, and the daughter nuclei move with a velocity v.
02

Analyze Kinetic Energy#

Analyze the process in terms of kinetic energy changes. We have the parent nucleus mass, M, and the change of mass, 螖m, during the process. Considering the mass-energy equivalence principle, we can convert the lost mass to energy using Einstein's famous equation: \(螖E = 螖m c^2\) Where 螖E is the change in energy, c is the speed of light, and 螖m is the mass change. This energy release will be converted to the kinetic energy of daughter nuclei.
03

Relate Kinetic Energy and Linear Momentum#

The kinetic energy of the daughter nuclei (K) can be expressed using momentum (p) and mass of the daughter nuclei (m): \(K = \frac{p^2}{2m}\) Where m is the mass of daughter nuclei, which is (M - 螖m). Now, we can equate the energy released with the kinetic energy of the daughter nuclei: \(螖E = K\) Replacing 螖E and K with their expressions: \(螖m c^2 = \frac{p^2}{2(M - 螖m)}\)
04

Solve for Velocity#

We have to find the speed, v, of daughter nuclei. The relationship between momentum and velocity is: \(p = m v\) where m is the mass of the daughter nuclei. Plug this expression into the previous equation: \(螖m c^2 = \frac{(m v)^2}{2(M - 螖m)}\) Now, we can solve this equation for velocity, v: \(v = c\sqrt{\frac{2螖m}{M - 螖m}}\)
05

Match Option with the Derived Formula#

Now, compare the derived formula with the given options: (A) \(c\sqrt{\frac{\Delta m}{M+\Delta m}}\) (B) \(c \frac{\Delta m}{M+\Delta m}\) (C) \(c \sqrt{\frac{2 \Delta m}{M}}\) (D) \(c \sqrt{\frac{\Delta m}{M}}\) The given formula in Step 4 is not matching with any of the given options. There might be a mistake, either in the question or in the analysis of the problem. The student should consult the teacher for clarification.

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Most popular questions from this chapter

When photons of energy \(5 \mathrm{eV}\) strike the surface of a metal \(A\), the ejected photoelectrons have maximum kinetic energy \(K_{A} \mathrm{eV}\) and de Broglie wavelength \(\lambda_{A}\). The maximum kinetic energy of photoelectrons liberated from another metal \(B\) by photons of energy \(5.30 \mathrm{eV}\) is \(K_{B}=\left(K_{A}-1.5\right) . \mathrm{eV}\). If the de Broglie wavelength of these photoelectrons is \(\lambda_{B}=2 \lambda_{A}\), then find \(K_{A}\) and \(K_{B}\).

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