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Chapter 5: Problem 20

taking \(h=6.63 \times 10^{-27} \mathrm{crgs}\) and \(c=300000 \mathrm{~km} \mathrm{~s}^{-1}\), calculate how many photons of wavelength \(5 \times 10^{-5} \mathrm{~cm}\) (in the yellowgreen of the visible spectrum) must fall per second on a blackened plate to produce a force of one dyne. [Answer: \(7.3 \times 10^{21}\). Hint: force equals momentum absorbed per unit time.]

Short Answer

Expert verified
Given Planck's constant \(h = 6.63 \times 10^{-27}\) crgs, and the speed of light \(c\). Answer: Approximately \(7.3 \times 10^{21}\) photons per second are required to produce a force of one dyne.

Step by step solution

01

Calculate momentum of one photon in the yellow-green spectrum

We are given the wavelength of the photon \(\lambda = 5 \times 10^{-5}\mathrm{~cm}\). Using the Planck's constant \(h = 6.63 \times 10^{-27}\mathrm{crgs}\), we can calculate the momentum of a single photon: \(p = \dfrac{h}{\lambda}\) Now, plug in the given values to find the momentum: \(p = \dfrac{6.63 \times 10^{-27}\mathrm{crgs}}{5 \times 10^{-5}\mathrm{~cm}} \approx 1.326 \times 10^{-22}\mathrm{~crg \cdot s \cdot cm^{-1}}\)
02

Calculate the number of photons required to produce the force

We are given the force to be 1 dyne. To find how many photons per second are required to produce the 1 dyne force, we need to divide the force by the momentum absorbed per second by each photon: Number of photons per second, \(N = \dfrac{\mathrm{Force}}{\mathrm{Momentum~per~Photon}}\) Plug in the values for force (1 dyne) and the calculated photon momentum: \(N = \dfrac{1\mathrm{~dyne}}{1.326 \times 10^{-22}\mathrm{~crg\cdot s\cdot cm^{-1}}}\)
03

Calculate the final result

Now, calculate the number of photons per second: \(N \approx \dfrac{1}{1.326 \times 10^{-22}} \approx 7.538 \times 10^{21}\) However, the number of photons should be in whole numbers. Therefore, we will round it up to the nearest whole number: \(N \approx 7.3 \times 10^{21}\) photons per second The required number of photons per second to produce a force of one dyne is approximately \(7.3 \times 10^{21}\).

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Most popular questions from this chapter

Prove that, in relativistic as in Newtonian mechanics, the time rate of change of the angular momentum \(h=\mathbf{r} \times p\) of a particle about an origin \(\mathrm{O}\) is equal to the couple \(\mathrm{r} \times \mathrm{f}\) of the applied force about \(\mathrm{O}\). If 1 \(^{\mu v}\) is the particle's four- angular momentum, and if we define the fourcouple \(G^{\mu v}=x^{\mu} F^{v}-x^{v} F^{\mu}\), where \(x^{\mu}\) and \(F^{\mu}\) are the four-vectors corresponding to \(\mathbf{r}\) and \(\mathbf{f}\), prove that \((\mathrm{d} / \mathrm{d} \tau) L^{\mu \nu}=G^{\mu \nu}\) and that the spacc-space part of this equation corresponds to the above threevector result.

In an inertial frame there is a uniform electric field e in the positive \(x\)-direction, which exerts a rest mass preserving force \(\mathbf{f}=q \mathbf{e}\) on a particle carrying a charge \(q\). If the particle is released from rest at the origin, and \(m, \tau\) denote its rest mass and proper time, prove that the particle will move (hyperbolically) according to the equation

The mass of a hydrogen atom is \(1.00814 \mathrm{amu}\), that of a neutron is \(1.00898\) amu, and that of a helium atom (two hydrogen atoms and two neutrons) is \(4.00388\) amu. Find the binding energy as a fraction of the total energy of a helium atom.

The position vector of the centre of mass of a system of particles in any inertial frame is defined by \(\mathbf{r}_{\mathrm{CM}}=\Sigma m r / \Sigma m\). If the particles suffer only collision forces, prove that \(\dot{\boldsymbol{r}}_{\mathrm{CM}}=u_{\mathrm{CM}}(\cdot \equiv \mathrm{d} / \mathrm{d} t)\); i.e. the centre of mass moves with the velocity of the CM frame. [Hint: \(\Sigma m\), \Sigmami are constant; \Sigmamir is zero between collisions, and at any collision we can factor out the r of the participating particles: \(r \Sigma \dot{m}=0 .]\)

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