复旦大学物理学系 2006~2007学年第二学期期末考试试卷 卷 课程名称:大学物理(下)课程代码:PHYS120002.05 开课院系:物理学系 考试形式:闭卷 姓名 学号 专业 题号 总分 得分 Electric constant 6o=8.85x10-12C2/(N.m2) Magnetic constant u0=4z×10N/A2=1.26×10°H/m Plank constant h=663x10-34.s=4.14x10er s Neutron mass 1675×10-kg 1. Derive an expression for the work required by an external agent to put the four charges together as indicated in the figure below. Each side of the square has
2. In the laboratory we produce magnetic fields using current-carrying wires rather than the motion of individual charges. To describe the magnetic field due to a current, we have the biot-Sanvart law B=s Consider the circuit of the figure below. The curved segments are arcs of circles of radii a and b: The straight segments are along the radii. find the magnetic field at P, assuming a current i in the circuit. (15%)
3.(a)We can write the electric and magnetic fields in the usual mathematical form of a sinusoidal traveling wave E(x, )=Em sin(kx-ar), B(x, s)=Bm sin(kx-aor) From Faraday's law, we have B Find the ratio of the amplitudes of the electric to the magnetic components of the
wave.(b)A vector called the Poynting vector is defined by s=E×B Please explain what this vector means. Note that it is a vector. (c)A radio station radiates a sinusoidal wave with an average total power of 50 kw. Assuming it radiates equally in all directions (which is unlikely in real-world situations), find the amplitudes of the electric and the magnetic components of the wave at a distance of 100 km from the antenna. (2%+4%+9%)
4.(a) Find the speed and the kinetic energy of a neutron having a de broglie wavelength of 0. 1 nm, typical of atomic spacing in crystals. (b) Label as true or false these statements involving the quantum numbers n, l, my. If a statement is false, write the correct one (1)The values of m that are allowed depend only on l and not on n(2) The n=4 shell contains four subshells. 3) The smallest value of n that can go with a i gIven Z is 1+1.(4)All states with H0 also have mo, regardless of the value of n (5) Every shell contains n subshells.(6)One of these subshells cannot exist: n=2, 1-1 n4, 1-3; n=3, 1-2; n=l, l-l.(c)For a hydrogen atom with F-l, the measured value of L, is 0. Find the possible values Lx can have.(6%+2%x6+2%) 5. Consider a grating with N slits, where the slit width a is assumed to be much smaller than the light wavelength. (a)Derive the intensity expression for the diffracted lights on the screen that is assumed far way from the grating. (b)Find the positions of the principal maxima of the diffraction pattern. (15%+5%)
6. You are required to solve one of the following two problems. Surely you could solve both, but no extra marks will be given 1).(a)As shown in the figure below, at the bottom of a pool with water (n=4/3)2.00 m deep, light rays emit upward in all directions. a circular area of light is formed at the surface of the water. Determine the radius r of the light circle.(b) Show that the light rays close to the normal appear to come from a point(2.00 /n=1. 5)m below the surface.(c) Suppose there is a camera that can take photos in the water. Now it is exactly beside the luminous body in question (a) at the bottom of the pool. A big mirror, whose surface parallel with the surface of the pool water, is directly over the pool surface by 4 m. The object of the camera is to take a photo of the image of the luminous boby formed by the mirror. Find how far away it will appear to the camera To solve this, you can use the result of question(b).(6%+5%+4%) (2). By making clever use of the correspondence principle, bohr calculated that the energies of the orbital quantum states of a hydrogen atom are given by E 882h2 n2 where m is the electron mass, e is the elementary charge, h is the Plank constant and n is a quantum number.(a)a neutral helium atom has two electrons. Suppose one of them is removed, leaving the helium ion He*. This ion, with its single electron, resembles the hydrogen atom, except that its nuclear charge is +2e rather than +e. Find the wavelength of the photon emitted when the electron makes a transition from the orbit n=5 to the orbit n=2.(b) The shell with n=l is called the K shell, that with n=2 the L shell that with n=3 the m shell and that with n=4 the N shell. If an electron falls from the L shell and moves in to fill a hole in the K shell we have the K, line if it falls from the M shell. we have K line. If an electron falls from the M shell to a hole in the l shell. we have the l line. if it falls from the N shell, we have Lg line. Let Aa and np lengths of the K i and Ap X-rays of an element. Show that the Bohr theory gives
and find the numerical value of k (c) Find the similar result for the relation between an
复旦大学物理学系 2006~2007学年第二学期期末考试试卷参考答案 A卷 课程名称:大学物理(下)课程代码:PHYS12000.05 开课院系:物理学系考试形式:闭卷 1 There are six interaction terms, one for every charge pair, number the charges clockwise from the upper left hand corner, then u2=9/rega u23=9" /4Eoa 34=92/4mE0a 2/4re0 43=q2/4xe0(2a) l24=q ld these terms and get u (4+√2) the amount of work required Is w=u 2 There are four current segments that could contribute to the magnetic field. the straight segments, however, contribute nothing because the straight segments carry currents either directly toward or directly away from the point p that leaves the two rounded segments. Each contribution to B can be found by idsxr rde The contribution to b from the top arc is: 4018 The direction is into the page
There is also a contribution from the above arc: 6io 2. the direction is out of 47b the page The net magnetic field at P is then: B=B+B,= The direction is out of the page.…… at Emesin(kx-ar)=-Bm. sin(kbx-aot E k=B C b k (b) The magnitude of the Poynting vector is the magnitude of the energy flow of the electromagnetic wave, it means the electromagnetic power per unit area in the space. The direction of the vector means where the energy of the wave is propogating It is also a function of the time which means it varies with the time. We more often than not pay attention to its average value First We consider the magnitude of the Poynting vector. We surround th antenna with an imaginary sphere of radius 100km. This sphere has area A=47R2=12.6×100m2 All the power radiated passes through this surface, so the power per unit area is s=p From the result of (a), one has e B E 10C Em=y2山cS 1.5 1.73×10-2N/C 0.5 B =E/
577×10-T h 6626*10j 4(1)D =3.96*103ms-1 Am(0.1*109m)(1675*1027kg) K=m2=(1.675*102kg)(3.96*103m·s-)2=1.31*102j=0.0818eV (2 )All of the six statements are true 2×6 (3)L1 Lx 5.(a). For every slit, the magnitude of the phasor is E1=E0 1,2,3, Then the total magnitude is E=∑E=E =E2e4“b+em+e2m+e?+…+c)m Eo kdN sin e kdN sin e E kd sin e sin 8 So the intensity reads kdN sin e kdN sin Ⅰ=I kdN sin e kd sin e (b). A principal maximam lies around the center where
