RIESGO IMPORTANTE RIESGO INTOLERABLE
1.9. NORMAS REFERENTES AL PERSONAL EN OBRA
According to Neil Bohr's hypothesis is the angular momentum of an electron is quantised.
mvr = n be taken into consideration.
c
Limiting line of Lyman series Lyman Series
(U.V. rays)
The maximum number of electrons that can be accommodated in an orbit is 2n2.
X-rays :
When fast moving electron strikes a hard metal, X-rays are produced. When the number of electrons striking the target metal increases, the intensity of X-rays increases. When the accelerating voltage/kinetic energy of electron increases λmin decreases. X-rays have the following properties :
(a) Radiations of short wavelength (0.01 Å – 10Å);
high pentrating power; having a speed of 3 × 108 m/s in vacuum.
λmin λ
Intensity Continuous spectrum (Varies & depends on accelerating voltage)
Kβ
Kα
Lγ Lβ
Lα
Characteristic spectrum (fixed for a target material)
(b) λmin = Isomers are nuclides which have identical atomic
number and mass number but differ in their energy states.
Atomic Structure, X-Ray & Radio Activity
P HYSICS F UNDAMENTAL F OR IIT-J EE
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
* The binding energy per nucleon is small for small nuclei.
* For 2 < A < 20, there are well defined maxima which indicate that these nuclei are more stable.
* For 30 < A < 120 the average B.E./A is 8.5 MeV / nucleon with a peak value of 8.8 MeV for Iron.
* For A > 120, there is a gradual decreases in B.E./nucleon.
* More the B.E./A, more is the stability.
Radioactivity :
β particles are electrons emitted from the nucleus.
(n → p + β)
(h) If a radioactive element decays by simultaneous emission of two particle then
dt
−dN
= λ1N + λ2N The following parameters remain conserved during a
nuclear reaction
(a) linear momentum (b) Angular momentum
(c) Number of nucleons (d) Charge
(e) The energy released in a nuclear reaction X + P → Y + Z + Q
Q = [mx + mp) – (my + mz)]c2 = ∆m × c2 Q = ∆m × 931 MeV
(f) In a nuclear fusion reaction small nuclei fuse to give big nuclei whereas in a nuclear fusion reaction a big nuclei breaks down.
Thermal neutrons produce fission in fissile nuclei.
Fast moving neutrons, when collide with atoms of comparable masses, transfer their kinetic energy to colliding particle and slow down.
According to Doppler's effect of light λ
1. The energy of an excited hydrogen atom is –3.4 eV.
Calculate the angular momentum of the electron according to Bohr theory.
Sol. The energy of the electron in the nth orbit is
2. The wavelength of the first member of the Balmer series in the hydrogen spectrum is 6563 Å. Calculate the wavelength of the first member of the Lyman series.
Sol. For the first member of the Balmer series
λ
For the first member of the Lyman series
´
3. Hydrogen atom in its ground state is excited by means of a monochromatic radiation of wavelength 970.6 Å. How many different wavelengths are possible in the resulting emission spectrum ? Find the longest wavelength amongst these.
Sol. Energy the radiation quantum E = hv = Energy of the excited sate
En = – 13.6 + 12.75 = – 0.85 eV
The number of possible transition in going to the ground state and hence the number of different wavelengths in the spectrum will be six as shown in the figure.
n 4 3
2
1
The longest wavelength corresponds to minimum energy difference, i.e., for the transition 4 → 3.
Now E3 = – 2
4. X-rays are produced in an X-ray tube by electrons accelerated through a potential difference of 50.0 kV.
An electron makes three collisions in the target before coming to rest and loses half its kinetic energy in each of the first two collisions. Determine the wavelengths of the resulting photons. Neglect the recoil of the heavy target atoms.
Sol. Initial kinetic energy of the electron = 50.0 keV Energy of the photon produced in the first collision, E1 = 50.0 – 25.0 = 25.0 keV
Wavelength of this photon λ1 =
Kinetic energy of the electron after third collision = 0 Energy of the photon produced in the third collision , E3 = 12.5 – 0 = 12.5 keV
This is same as E2. Therefore, wavelength of this photon, λ3 = λ2 = 0.99 Å
5. In an experiment on two radioactive isotopes of an elements (which do not decay into each other), their mass ratio at a given instant was found to be 3. The rapidly decaying isotopes has larger mass and an activity of 1.0 µCi initially. The half lives of the two isotopes are known to be 12 hours and 16 hours.
What would be the activity of each isotope and their mass ratio after two days ?
Sol. We have, after two days, i.e., 48 hours,
Qualitative Analysis :
Qualitative analysis of an organic compound involves the detection of various elements present in it. The elements commonly present in organic compounds are carbon, hydrogen, oxygen, nitrogen, halogens, sulphur and sometimes phosphorus.
Detection of carbon and Hydrogen :
Principle. Carbon and hydrogen are detected by strongly heating the organic compound with cupric oxide, (CuO). The carbon present in the organic compound is oxidised to carbon dioxide and hydrogen is oxidised to water. Carbon dioxide is tested by lime water test, whereas water is tested by anhydrous copper sulphate test.
Mixture of orgainc compoud and dry copper oxide (CuO)
Cotton plug Anhydrous
copper sulphate (white)
Guard tube containing sodalime
Lime water
Reactions :
C + 2CuO → CO2 + 2Cu in the compound
2H + CuO → H2O + Cu in the compound
CO2 + Ca(OH)2 → CaCO3 + H2O limewater milky
5H2O + CuSO4 (anhyd) → CuSO4.5H2O
white blue
Process : The given organic compound is mixed with dry cupric oxide (CuO) and heated in a hard glass tube. The products of the reaction are passed over (white) anhydrous copper sulphate and then bubbled through limewater. The copper sulphate turns blue (due to the formation of CuSO4.5H2O) by water vapour, showing that the compound contains hydrogen. The limewater is turned milky by CO2, showing that the compound contains carbon.
Detection of Nitrogen, Sulphur and Halogens :
Nitrogen, sulphur and halogens in any organic compound are detected by Lassaigne's test.
Preparation of Lassaigne's Extract (or sodium extract):
A small piece of sodium is gently heated in an ignition tube till it melts. The ignition tube is removed from the flame, about 50–60 mg of the organic compound added and the tube heated strongly for 2–3 minutes to fuse the material inside it.
After cooling , the tube is carefully broken in a china dish containing about 20–30 mL of distilled water.
The fused material along with the pieces of ignition tube are crushed with the help of a glass rod and the contents of the china dish are boiled for a few minutes. The sodium salts formed in the above reactions (i.e., NaCN, Na2S, NaX or NaSCN) dissolve in water. Excess of sodium, if any, reacts with water to give sodium hydroxide. This alkaline solution is called Lassaigne's extract or sodium extract. The solution is then filtered to remove the insoluble materials and the filtrate is used for making the tests for nitrogen, sulphur and halogens.
Reactions : An organic compound containing C, H, N, S, halogens when fused with sodium metal gives the following reactions.
C + N + Na →fusion NaCN in organic compound sodium cyanide
X(Cl, Br, I) + Na fusion → NaX(X=Cl,Br, I) from organic compound sodium halide
S + 2Na fusion → Na2S from organic compound sodium sulphide If nitrogen and sulphur both are present in any organic compound, sodium thiocyanate (NaSCN) is formed during fusion which in the presence of excess sodium, forms sodium cyanide and sodium sulphide.
Na + C + N + S fusion → NaCNS in organic compound sodium thiocyanate Detection of Nitrogen :
Take a small quantity of the sodium extract in a test tube. If not alkaline, make it alkaline by adding 2–3 drops of sodium hydroxide (NaOH) solution. To this solution, add 1 mL of freshly prepared solution of ferrous sulphate. Heat the mixture of the two solutions to boiling and then acidify it with dilute sulphuric acid. The appearance of prussion blue or green colouration or precipitate confirms the presence of nitrogen in the given organic compound.