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- 1. V R P 1 2 2 3 10 120 3 2. V R 1 30 90 3 10 3. in V R P 1 10 3 2 2 3 T 40 co2/3 2/3 1/2 1/2 1/2 According to the plots choose the appropriate indices from below: A. [120] B. [001] C. [10] D. [112] E. [301] F. (111] G. H. [012 1. (230) J. (233) K. (131) L (135) M. (300) N. (100) O. (111) P. (010) What is the Direction of OA?49
- 7. Calculate the energy (J) of a pure elemental sample weighing 0.5 g. 8. Determine the mass (g) of a sample with an energy of 1.78 eV.4. A particular atom has two energy levels with a transition wavelength of 420 nm. At 297 K there are 2.5 1021 atoms in the lower state. (a) How many atoms are in the upper state? (b) Suppose that 1.8 1021 of the atoms in the lower state are pumped to the upper state. How much energy could this system release in a single laser pulse?1. List all possible states of a hydrogen atom that have E=-1.51 eV.
- 1. Using the equation p-n+ N – N = 0 and assuming ND = Np and N4 = N4, calculate and plot (a) n versus T , (b) n versus 1/T, and (c) Ef-E¡ versus T for Si for 250 KIn the equation below, the Balmer series involves the emission lines (wavelengths) obtained when electrons go from higher energy (excited) level to 1 1 R n' 1 where, R = 1.097 x 107 m-1 %3D -- A. The ground level (n = 1) B. The first atomic level (n = 2) C. The third atomic level (n = 3) D. The fourth atomic level (n = 4) E. The fifth atomic level (n = 4) %3D %3DQUESTION 10 Which of these expressions would yield the wavelength of light in meters emitted when an electron drops from orbit n = 3 to n = 2 in a Bohr hydrogen atom? Given h = 4.14 x 10-15 eVs and c = 3.00 x 108 m/s. a. hc/1.89 b. 1.89 x h x c c. 1.89/hxc d. (1.51 + 3.4)/hc e. hc/3.4Which of these expressions would yield the wavelength of light in meters emitted when an electron drops from orbit n = 3 to n = 2 in a Bohr hydrogen atom? Given h = 4.14 x 10-15 eVs and c = 3.00 x 108 m/s. a. 1.89/hxc b. hc/1.89 c. 1.89 x h x c d. (1.51 + 3.4)/hc e. hc/3.44. Show that the internal energy, U, of an Einstein crystal is given by: 3Nhv 3Nhvexp(-ßhv) 2 1-exp(-Bhv) + U = Hint: The following identity will avoid you having to use the chain rule: d In(1− ef(x)) = −(ƒ'(x)) e²“ dx 1-eIf you are shining light on a Pt electrode and no electrons are emitted, what do you have to do to generate electrons? Decrease the wavelength of the light to increase the energy of the photons to emit electrons Electrons can not be emitted from Pt because like any other noble metal it does not react, even under light exposure (unlike other metals, e.g. Na, Sr or Cu) Increase the wavelength of the light to increase the energy of the photons to emit electrons Wait longer times or increase the intensity of the light source, since the energy of the absorbed light is not high enough to emit electrons from Pt surfaceSEE MORE QUESTIONS