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Math problems - Essay Example

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1a) i. The lowest level in the hydrogen atom is level two n=2 which is involved in all transitions for the absorption lines in the visible region of the spectrum. ii. -1.0826 eV is required to ionize hydrogen from this level and te formula is given as E = -R/n^2, where R = Rydberg's (energy) constant = 13.598 eV*, and n = shell number that the electron is in…
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Math problems
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Download file to see previous pages ii) -1.89 lie in IR region. 1c)Because its closer to the nucleus and having some effect on the other electron present in the higher shell Q2 a) Power output of sun = 3.8 *1026W Radius = 1.4 * 108 Power density = ? V= 4/3 ?r3 V= 4/3 ? (1.4*108)3 =1.15*1025 Power density: Power/volume = 3.8*1026/1.15*1025 1ev= 1.6*10-19J Total energy released for He nucleus is 26.7 Mev 26.7*106*1.6*10-19 4.27*10-12 energy is released for 1 he atom =>3.8*1026 J produce 3.8*1026/4.27*10-12 *1/(4*?*(1.4*108)3/3) Helium atoms per second per cubic meter (here assume time =1sec therefore energy =3.8 *1026W *1sec 7.74*1012 Helium atoms per second per cubic meter 2b) In He both the nucleons are there means protons and neutron,2 protons and 1 neutron is there. In first equation there are less number of nucleons involve so there nuclear force will be less and thus it will be more reactive there fore they are less stable while on the other hand when the nucleons are more in number as it is in step 2 therefore there will be strong nuclear force, will be less reactive and more stable. 2c) When the fusion reaction occurs so at that time two atoms combine and produce larger atom and release high energy in the form of binding energy of nucleons. As this process continues till the formation of iron Fe 56,so at that time binding energy of electron is minimum that is most negative and now if the more heavier atom is required to be formed so more energy will release Q3 a i) Data: Redshift = z = 0.13 Speed of light =c= 3 * 108 ms-1 Hubble constant = H0 = 70 kms-1Mpc-1 Distance to the galaxy = r=? Formula: 1. H0 = v / r Here v = apparent speed of galaxy 2. v = z * c Solution: v = z * c =0.13 * 3 * 108 v = 3.9 * 107 ms-1 v = 3.9*107*10-3kms-1 v = 3.9*104 kms-1 H0 = v / r r= v / H0 r = 3.9*104 /70 r = 5.57*102 Mpc Q3 a ii) Data Red shift =z = 0.13 ?0=589nm ?1=? Formula ??=?1-?0 z=??/?0 Solution z= ??/?0 0.13= ??/589 ??=76.57nm ?1-?0 =76.57 ?1 =76.57+589 ?1=665.57nm ?1=6.65*102 nm Q3 b i) As the wavelength of hypothetical object is different as compare to the wavelength of the objects which are already present in the cluster and this wavelength is very large which causes this hypothetical object to move out of the galaxy that’s why that this object is not part of this cluster, and is actually more distant. Q3 b ii) The answer is not in the book. Or no relative material is in the book kindly search yourself Q#4 General relativity and quantum gravity depart from Newton’s theory. The gravitational force of attraction is described by Newton’s law of gravity. Einstein’s theory of general relativity describes the interaction between space and the matter within it. When the masses become very large, this theory provides a more accurate description of gravity than does Newton’s law. General relativity also predicts the existence of gravitational radiation, which is emitted by massive objects that undergo an acceleration. There is good evidence that such radiation is being generated by binary pulsars. A convincing theory of quantum gravity has yet to be formulated, but it will involve quanta referred to as gravitons which interact with everything. Einstein’s theory of general relativity reproduced all the old results of Newton, but without even using the idea of weight. The core of general relativity is the interaction between ‘space’ ...Download file to see next pagesRead More
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