1. In the Bohr model, compare the magnitudes of the electron s kinetic and potential energies in orbit. What does this imply?


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1 Assignment 3: Bohr s model nd lser fundmentls 1. In the Bohr model, compre the mgnitudes of the electron s kinetic nd potentil energies in orit. Wht does this imply? When n electron moves in n orit, the Coulom force cting on it lnces the centripetl ccelertion, which keeps the electron in its orit. i.e., 1 Ze 2 4πε 0 r 2 1 Ze 2 4πε 0 r mv2 r mv 2, (1 where v is the speed of electron nd r is the rdius of the orit. The kinetic energy of this electron is, K 1 2 mv2 1 ( 1 Ze 2 2 4πε 0 r (using eqution1 where U Ze2 4πε 0 r 1 2 (U K 1 2 U. is the potentil energy of this electron. This implies tht the mgnitude of the kinetic energy of n electron is hlf the mgnitude of potentil energy. Now, the totl energy of this electron is, E K + U 1 2 U + U 1 2 U, which immeditely implies tht if the potentil energy is negtive, the totl energy will lso e negtive, which is exctly the cse in hydrogen tom. 2. The sorption spectrum of n element shows fewer lines thn its emission spectrum. Could you reson why? 1
2 When we het n oject to otin n emission spectrum, the spectrum consists of spectrl lines of severl frequencies in the electromgnetic spectrum. But in the cse of sorption, some lines re missing or hve reduced intensities. The reson is tht in sorption spectroscopy, the tom strts off in the Boltzmnn distriution, n exp E i /(k B T, which is the ground stte. Not ll upper sttes will e populted s result of the incident rdition. 3. In strs, the Pickering series is found in the He + spectrum. It is emitted when the electron in He + jumps from higher levels into the level with n 4. ( Stte the exct formul for the wvelength of lines elonging to this series. ( In wht region of the spectrum is the series? (c Find the wvelength of the series limit. (d Find the ioniztion potentil, if He + is in the ground stte, in electron volts. When n electron mkes trnsition from n initil quntum stte (n i to finl quntum stte (n f, the frequency f of the emitted electromgnetic rdition is, f E i E f h ( 2 ( 1 mz 2 e 4 1 4πε 0 4π 3 n 2 f 1 n 2 i where E n me4 Z 2 (4πε n 2 In terms of reciprocl wvelength κ 1 f, we hve, λ c ( 2 ( 1 mz 2 e 4 1 κ 1 4πε 0 4π 3 c n 2 f n 2 i R Z 2 ( 1 n 2 f 1 n 2 i ( 2 1 me The Ryderg constnt, R 4, is equl to πε 0 4π 3 c m 1. In the cse of Pickering series found in He + spectrum, the electron jumps from higher 2
3 levels into n f 4. Therefore, ( 1 1 λ R Z 2 n 2 f ( 1 R Z 2 λ 1 n 2 i 16 1 n 2 i R Z 2 ( n 2 i 16 16n 2 i 16n2 i 1 n 2 i 16 R Z 2 ( If we tke n i 5, i.e., the trnsition is tking plce from n 5 to n 4 energy level, the corresponding wvelength is, which is in the ultrviolet region. λ 16n2 i 1 n 2 i 16 R Z nm, (c The Pickering series ctully corresponds to the trnsitions from n 5, n 7, nd n 9 to n 4. Therefore, the wvelength of series limit corresponds to the trnsition from n 9 to n 4, for which, which is in the ultrviolet region. λ 16n2 i 1 n 2 i 16 R Z nm, (d The ioniztion potentil of helium (He + hving Z 2 t ground stte (n 1 is, me 4 Z 2 1 (4πε n ev n ev. 4. A muon is prticle with chrge equl to tht of n electron nd mss equl to 207 times the mss of n electron. Muonic led is formed when 208 P cptures muon to 3
4 replce n electron. Assume tht the muon moves in such smll orit tht it sees nucler chrge of Z 82. According to the Bohr theory, wht re the rdius nd energy of the ground stte of muonic led? The effective mss of muon, when it moves in smll orit nd sees nucler chrge of Z 82, is given y the following expression, µ m µm m µ + M, where m µ 207m e is the mss of muon nd M 82m p is the mss of nucleus tht it sees. Therefore, µ (207 m e(82 m p 207 m e + 82 m p ( kg( kg kg + ( kg kg. According to Bohr theory, the rdius is, n 2 2 r 4πε 0 µze ( ( ( m. The energy of the ground stte of muonic led is, E n me4 Z 2 (4πε n 2 ( ( ( ( ( MeV. 5. If the electronic structure of the H tom is quntized, why do we get continuous emission spectrum from the H on the sun? On the solr surfce, the temperture is very high leding to high temperture plsm of hydrogen which possesses continuous rnge of trnsltionl kinetic energies. Hence, this hot plsm cn e likened to lckody possessing continuous spectrum. 4
5 6. An tom hs two energy levels with trnsition wvelength of 5800 Å. At room temperture toms re in the lower stte. ( How mny occupy the upper stte, under conditions of therml equilirium? ( Suppose insted tht toms re pumped into the upper stte, with in the lower stte. How much energy in joules could e relesed in single pulse? If N is the higher energy stte nd N is the lower energy stte, then the energy difference etween these two sttes cn e clculted using trnsition wvelength of 5800 Å. i.e., E E E E hf hc λ ( Jsec( m/sec m J k B T J/K 300 K J. According to the Boltzmnn distriution, the popultion in ny stte i is, N i e E i/k B T. Therefore, the rtio of N nd N is, N eq N eq e (E E /k B T e ( J/ J N eq N eq ( ( , which is very smll. This shows tht the energy gp etween the given sttes N nd N, t room temperture, is so lrge tht lmost ll the electrons re present in lower stte N. There is hrdly ny electrons in the upper stte. 5
6 ( When toms re pumped in the upper stte, this is nonequilirium sitution. Energy is relesed up till equilirium is restored. Therefore, We re interested in numer of toms which restores the equilirium. Since the rtio, N /N , remins the sme, therefore, ( The numer of toms which contriute to restore equilirium is, N The energy relesed in single pulse is thus, E N hc λ J 240 J. 7. Determine pproximtely the rtio of the proility of spontneous emission to the proility of stimulted emission t room temperture in ( the Xry region of the electromgnetic spectrum, ( the visile region. 6
7 ( The rtio of the Einstein A nd B coefficients is, 8πhf 3 B 21 c 3 f c λ m/sec m /sec 8π( J sec( /sec 3 B 21 ( m/sec J sec m 3. This shows tht the spontneous emission overwhelms the stimulted emission. Therefore, it is very difficult to mke lsers with Xrys. ( For the visile region of electromgnetic spectrum, we hve 8πhf 3 B 21 c 3 f c λ m/sec m /sec 8π( J sec( /sec 3 B 21 ( m/sec J sec m 3. This shows tht the stimulted emission domintes over the spontneous emission, fvoring the lsing ction. 7
, and the number of electrons is 19. e e 1.60 10 C. The negatively charged electrons move in the direction opposite to the conventional current flow.
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