320 + Interview Questions in NUCLEI IN GENERAL AWARENESS Page 6 InterviewSolution

251.	Beta rays emitted by a radicactive material areA. Electromagnetic radiationB. Neutral particlesC. The electrons orbiting around the nucleusD. Charged particles emitted by nucleus
Answer» Correct Answer - D

Discussion

252.	When a deuterium is bombarded on `._(8)O^(16)` nucleus, an `alpha`-particle is emitted, then the product nucleus isA. `._7N^13`B. `._5B^10`C. `._4Be^9`D. `._7N^14`
Answer» Correct Answer - D

Discussion

253.	Boron rods are used in nuclear reactors becauseA. boron can absorb neutronsB. strength is given to the plantC. chain reaction of neutrons and `""_(92)U^(240)`D. the reactors look good
Answer» Correct Answer - A

Discussion

254.	The half life of `._92U^(238)` against `alpha` -decay is `4.5xx10^9` years. What is the activity of 1g sample of `._92U^(238)`?
Answer» `T_(1//2)=4.5xx10^(9)y` `=4.5xx10^(9)yxx3.16xx10^(7)s//y` `=1.42xx10^(17)s`One k mol of any isotope contains Avogadro’s number of atoms, and so 1g of `._(92)238)U` contains `(1)/(238xx10^(-3)kmolxx6.025xx10^(26)` atoms /Kmol `=25.3xx10^(20)` atoms. The decay rate R is R = λN `=(0.693)/(T_(1//2)N=(0.693xx25.3xx10^(20))/(1.42xx10^(17))s^(-1)` `=1.23xx10^(4)s^(-1)` `=1.23xx10^(4)Bq`

Discussion

255.	The limit of resolution of a microscope is the least distance between two point objects which can be seen clearly and distinctively with it. For a naked eye, limit of resolution is about `10^(-4)` m or 0.1 mm. The limit of resolution for a microscope is of the order of `lambda/2` , where `lambda` is wavelength of light used. For an optical microscope , using visible light, limit of resolution is `"500 nm"/2`=250 mm. This is 400 times smaller than naked eye. So the useful magnification produced by it is 400. Smaller is the limit of resolution, greater is the magnification that can be achieved. By using shorter wavelengths , we can improve resolution. For examination of a microorganism, much higher limit of resolution is needed, the Wave nature of electron provides a means for probes of very small size organisms. An electron beam accelerated by a high potential difference possess a very short wavelength. In an electron microscope magnetic lenses are used to control the path of electrons. The image of object is obtained on a fluorescent screen or on a photographic plate. The electron microscope with its high magnifying power and resolving power , is one of the most indispensable and powerful tool for research in science , medicine and industry. The advantage of electron microscope over optical microscope is thatA. It gives images which can be directly seen with naked eyeB. It gives a better resolution and magnificationC. It is very simple in its workingD. It uses magnetic lens in place of glass lens
Answer» Correct Answer - B

Discussion

256.	The limit of resolution of a microscope is the least distance between two point objects which can be seen clearly and distinctively with it. For a naked eye, limit of resolution is about `10^(-4)` m or 0.1 mm. The limit of resolution for a microscope is of the order of `lambda/2` , where `lambda` is wavelength of light used. For an optical microscope , using visible light, limit of resolution is `"500 nm"/2`=250 mm. This is 400 times smaller than naked eye. So the useful magnification produced by it is 400. Smaller is the limit of resolution, greater is the magnification that can be achieved. By using shorter wavelengths , we can improve resolution. For examination of a microorganism, much higher limit of resolution is needed, the Wave nature of electron provides a means for probes of very small size organisms. An electron beam accelerated by a high potential difference possess a very short wavelength. In an electron microscope magnetic lenses are used to control the path of electrons. The image of object is obtained on a fluorescent screen or on a photographic plate. The electron microscope with its high magnifying power and resolving power , is one of the most indispensable and powerful tool for research in science , medicine and industry. The working of an electron microscope uses properties of electrons related toA. Only wave nature of electronB. Only particle nature of electronC. Both particle nature as well as wave natureD. None of these
Answer» Correct Answer - C

Discussion

257.	After 300 days, the activity of a radioactive sample is 5000 dps (disintegrations per sec). The activity becomes 2500 dps after another 150 days. The initial activity of the sample in dps isA. 20000B. 10000C. 7000D. 25000
Answer» Correct Answer - A As, activity droped from 5000 dps to 2500 dps in 150 days means `T_(1//2)=150"days"` `implies300"days"=2T_(1//2)` So, initial activity `2xx5000xx2=20000` dps

Discussion

258.	In the given reaction `._z X^A rarr ._(z+1)Y^A rarr ._(z-1) K^(A - 4) rarr ._(z-1) K^(A - 4)` Radioactive radiations are emitted in the sequence.A. `alpha,betagamma`B. `gamma,alpha,beta`C. `beta,alpha,gamma`D. `lamda,beta,alpha`
Answer» Correct Answer - C Suppose a given nucleus is `""_(Z)X^(A)`. In `alpha-"decay"`, the mass number of the product nucleus I sfour less than that of decaying nucleus, while the atomic number decreases by two. `""_(Z)X^(A)overset(alpha-"decay")to""_(Z-2)X^(A-4)""_(2)He^(4)(alpha)` In `beta-"decay`, the mass number of product nucleus remains same but atomic number increases or decreases by one. `""_(Z)^(A)Xoverset(beta-"decay")to""_(Z)^(A)+""_(1)Y+_(-1)^(0)e+v` or `""_(Z)^(A)Xoverset(beta-"decay")to""_(Z)^(A)+""_(1)Y+_(+1)^(0)e+v` Gamma decay is the phenomenon of emission of gamma rays photon from a radioactive nucleus. `""_(Z)X^(A)to""_(Z)X^(A)+y` Hence, the given reaction will have sequence as follows `""_(Z)X^(A)overset(beta-"emission")to""_(Z+1)Yoverset(alpha-"emission")to` `""_(Z-1)K^(A-Y)overset(y-"emission")to""(Z-1)K^(A-y)`

Discussion

259.	In the given reaction `._z X^A rarr ._(z+1)Y^A rarr ._(z-1) K^(A - 4) rarr ._(z-1) K^(A - 4)` Radioactive radiations are emitted in the sequence.A. `alpha,beta,gamma`B. `gamma,alpha,beta`C. `beta,alpha,gamma`D. `beta,gamma,alpha`
Answer» Correct Answer - C

Discussion

260.	A radioactive element X with half life 2 h decays giving a stable element Y. After a time t, ratio of X and Y atoms is 1:16 .Time t isA. 6 hB. 4 hC. 8 hD. 16 h
Answer» Correct Answer - c `N/N_0=(1/2)^n=1/16` `therefore` n=4 `t=nT_(1//2)=4xx2`=8 h

Discussion

261.	What fissionable material was used in the bomb dropped at Nagasaki (Japan) in the year 1945 ?A. UraniumB. NeptuniumC. BerkeliumD. Plutonium
Answer» Correct Answer - D

Discussion

262.	Write the truth table for the circuit shown in figure given below. Name the gate that the circuit resembles.
Answer» The circuit resemble AND gate. The boolean expression of this circuit is, `V_(0) = A.B " i.e., " V_(0)` equals A AND B. The truth table of this gate is as given below `{:(ul(A" "B" "V_(0) = A.B)),(0" "0" "0),(0" "1" "0),(1" "0" "0),(ul(1" "1" "1)):}`

Discussion

263.	During a negative beta decay,A. an atomic electron is ejectedB. an electron which is already present with in the nucleus is ejectedC. a neutron in the nucleus decays emitting on electronD. a part of the binding energy is converted into electron
Answer» Correct Answer - A Negative `Beta`-decay is expressed by the equation, `nto p^(+)e^(-)+v^(-).`

Discussion

264.	A radioactive nucleus `._92 X^235` decays to `._91 Y^231`. Which of following particles are emitted ?A. One alpha and one electronB. Two deuterons and one positronC. One alpha and one protonD. One proton and four neutrons
Answer» Correct Answer - A `""_(92)X^(235)overset(alpha)to_(90)X^(231)overset(beta^(-))to_(91)Y^(231)` Therefore, one alpha and one electron are emitted.

Discussion

265.	Assertion:An `alpha`-particle is emitted when uranium 238 decays into thorium Reason : The decay of uranium 238 to thorium is repesented by `._92^238 U to ._90^234Th + ._2^4 He `. The helium nuclei is called an alpha particle.A. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false
Answer» Correct Answer - a

Discussion

266.	Uranium ores on the earth at the present time typically have a composition consisting of `99.3%` of the isotope `._92U^238` and `0.7%` of the isotope `._92U^235`. The half-lives of these isotopes are `4.47xx10^9yr` and `7.04xx10^8yr`, respectively. If these isotopes were equally abundant when the earth was formed, estimate the age of the earth.
Answer» Let `N_(0)` be number of atoms of each isotope at the time of formation of the earth `(t=0)` and `N_(1)` and `N_(2)` the number of atoms at present `(t=t)`. Then, according to law of radioactivity, ` N_(1)=N_(0)e^(-lamda_(1) t)" "`…(i) and `" "N_(2)=N_(0)e^(-lamda_(2)t)" "...(ii)` `therefore" "(N_(1))/(N_(2))=e^((lamda_(2)-lamda_(1))t )" "`...(iii) Further it is given that, `(N_(1))/(N_(2))=(99.3)/(0.7)" "`...(iv) Equating Eqs. (iii) and (iv) and taking log on both sides, we have `" "(lamda_(2)-lamda_(1))t=In((99.3)/(0.7))` `therefore" "t=((1)/(lamda_(2)-lamda_(1)))In ((99.3)/(0.7))` Where `lamda_(1) and lamda_(2)` are the decay constant. Substituting the values, we have `" "t=(1)/((0.693)/(7.04xx10^(8))-(0.693)/(4.47xx10^(9)))In((99.3)/(0.7))` `" "(becauset_(1//2)(0.693)/(lamda)rArrlamda=(0.693)/(t_(1//2)))` or age of the earth, `t=5.97xx10^(9)` yr

Discussion

267.	Calculate the binding energy and binding energy per nucleon in MeV for carbon-12 nucleus. Given that mass of the proton is 1.00727 amu while the mass of the neutron is 1.00866 amu.
Answer» Mass defect ΔM = Zm_P + (A - Z)m_N - M For carbon -12 nucleus Z = 6, A = 12 and M = 12 amu => ΔM = 6 x 1.00727 + 6 x 1.00866 - 12 = 0.09558 amu E = ΔM x 931 MeV = 0.09558 x 931 = 88.985 MeV

267.

Calculate the binding energy and binding energy per nucleon in MeV for carbon-12 nucleus. Given that mass of the proton is 1.00727 amu while the mass of the neutron is 1.00866 amu.

Answer»

Mass defect

ΔM = Zm_P + (A - Z)m_N - M

For carbon -12 nucleus

Z = 6, A = 12 and M = 12 amu

=> ΔM = 6 x 1.00727 + 6 x 1.00866 - 12 = 0.09558 amu

E = ΔM x 931 MeV = 0.09558 x 931 = 88.985 MeV

Discussion

268.	What is the binding energy per nucleon of `_(6)C^(12)`nucleus? Given , mass of `C^(12)` `(m_(c))_(m)` = 12.000 u Mass of proton `m_(p)` = 1.0078 u Mass of neutron `m_(n)` = 1.0087 u and 1 amu = 931.4 MeVA. 5.26 MeVB. 6.2 MeVC. 4.65 MeVD. 7.68 MeV
Answer» Correct Answer - D Binding energy per nucleon `([{m_(p)Z+m_(n)(A-Z)}-M]xx931)/("Mass number A")` `=((6xx1.0078+6xx1.0087-12)/(12))xx931.4` = 7.68 MeV

Discussion

269.	Consider the circuit arrangnent shown in Fig. for studying input and output characteristics of npn transistor in CE configuration. Select the values of `R_` and `R_` for a transistor whose `V_(BE)=0.7V`, so that the transistor is operating at point Q as shown in the characteristics shown in Fig. Given that the input impedance of the transistor is very small and `V_(C C)=V_(BB)=16V`, also find the voltage gain and power gain of circuit making appropriate assumptions.
Answer» Given, `V_(BE) = 0.7 V, V_(C C) = V_(BB) = 16 V` `V_(CE) = 8V` `I_(C) = 4 mA = 4 xx 10^(-3) A` `I_(B) = 30 mu A = 30 xx 10^(-6) A` For the ouput characteristic at `theta`, `V_(C C) = I_(C) R_(C) + V_(CE)` `R_(C) = (V_(C C) - V_(CE))/(I_(C)) = (16 - 8)/(4 xx 10^(-3)) = (8 xx 1000)/(4) = 2 k Omega` Using the relation, `V_(BB) = I_(B) R_(B) + V_(BE)` `R_(B) = (V_(BB) - V_(BE))/(I_(B)) = (16 - 0.7)/(30 xx 10^(-6))` `= 150 xx 10^(3) Omega = 510 k Omega` `beta = (I_(C))/(I_(B)) = (4 xx 10^(-3))/(30 xx 10^(-6)) = 133` Voltage gain `= beta (R_(C))/(R_(B)) = (133 xx 2 xx 10^(3))/(510 xx 10^(3)) = 0.52` Power gain `beta xx` Voltage gain `= 133 xx 0.52 = 69`

Discussion

270.	Define Becquerel and Curie.
Answer» Becquerel : 1 disintegration or decay per second is called Becquerel. It is SI unit of activity. i.e., Becquerel `= ("1 disintegration or decay")/("second")` Curie : `3.7xx10^(10)` decays per second is called Curie. 1 Curie `= 1Ci=(3.7xx10^(10)" decays")/("second")=3.7xx10^(10)` Bq.

Discussion

271.	What is the role of controlling rods in a nuclear reactor ?
Answer» In nuclear reactor controlling rods are used to absorb the neutrons. Cadmium, boron materials are used in the form of rods in reactor. These control the fission rate.

Discussion

272.	Why are nuclear fusion reactions called thermo nuclear reactions ?
Answer» Nuclear fusion occurs at very high temperatures. So it is called as thermo nuclear reaction.

Discussion

273.	Which of the following equactions pick out the possible nuclear fusion reactions?A. `._6^13C +._1^1H to ._6^14C` +4.3 MeVB. `._6^12C + ._1^1H to ._7^13N` + 2 MeVC. `._7^14N + ._1^1H to ._8^15O ` + 7.3 MeVD. `._92^235C + ._0^1N to ._54^140 Xe + ._38^94Sr + ._0^1n + ._0^1n` + 200 MeV
Answer» Correct Answer - a During nuclear fusion , two or more light nuclei combine to form a heavier and more stable nucleus . As `._6C^14` is radioactive , so (a) is not possible .

Discussion

274.	`A U^(235)` reactor generated power at a rate of P producting `2xx10^(18)` fussion per second. The energy released per fission is 185 MeV. The value of P isA. `59.2` MWB. `370xx10^(18)MW`C. `0.59 MW`D. 370 MW
Answer» Correct Answer - A Power, `P=(nE)/(t)` Given, `n=2xx10^(18)` fission per second E=185MeV Here, `P=(2xx10^(18)xx185xx10^(6)eV)/(1)` `implies=2xx10^(18)xx185xx10^(6)xx16xx10^(-19)=59.2MW`

Discussion

275.	If `200 MeV` energy is released in the fission of a single `U^235` nucleus, the number of fissions required per second to produce `1` kilowatt power shall be (Given `1 eV = 1.6 xx 10^-19 J`).A. `3.125xx10^13`B. `1.52xx10^6`C. `3.125xx10^12`D. `3.125xx10^14`
Answer» Correct Answer - a Let the number of fussions per second be n. Energy released per second = n x 200 MeV = n x 200 x 1.6 x `10^(-13)` J Energy required per second = power x time =1 kW x 1s = 1000 J `therefore n xx 200 xx 1.6 xx 10^(-13)`=1000 or `n=1000/(3.2xx10^(-11)) =10/3.2 xx 10^13 = 3.125xx10^13`

Discussion

276.	In a nuclear reactor, the number of `U^(235)` nuclei undergoing fissions per second is `4xx10^(20).` If the energy releases per fission is 250 MeV, then the total energy released in 10 h is `(1 eV= 1.6xx10^(-19)J)`A. `576xx10^(6)J`B. `576xx10^(12)J`C. `576xx10^(15)J`D. `576xx10^(12)J`
Answer» Correct Answer - B (B) the fission per second `=4xx10^(20)` So , the energy relased per second `=4xx10^(20) xx250MeV ` `=4xx10^(20) xx250xx10^(6) xx1.6xx10^(-19) J` therefore , energy relased in 10 h`= 36xx10^(3) ` s is `E=36xx10^(3) xx4xx10^(20) xx250xx10^(6) xx1.6xx10^(-19) J` `=36xx4xx250xx1.6xx10^(10) J= 57600xx10^(10)` `=576xx10^(12)J`

Discussion

277.	Complete the following nuclear reactions :(i) 20884Po → 20482Pb + ........(ii) 3215P → 3216S + ..........
Answer» We have (i) ²⁰⁸₈₄Po → ²⁰⁴₈₂Pb + ⁴₂He + Q (Also accept if Q is not written) (ii) ³²₁₅P → ³²₁₆S + ⁰_-1e + v [Also accept if r.r is not written]

277.

Complete the following nuclear reactions :(i) 20884Po → 20482Pb + ........(ii) 3215P → 3216S + ..........

Answer»

We have

(i) ²⁰⁸₈₄Po → ²⁰⁴₈₂Pb + ⁴₂He + Q

(Also accept if Q is not written)

(ii) ³²₁₅P → ³²₁₆S + ⁰_-1e + v

[Also accept if r.r is not written]

Discussion

278.	Complete the following nuclear reactions :(a)10 5B + 01n → 24He + ........(b) 9442Mo + 21n → 9543Te + .........
Answer» (a)¹⁰₅B + ₀¹n → ₂⁴He + ⁷₃Li (b) ⁹⁴₄₂Mo + ²₁n → ⁹⁵₄₃Te + ¹₀n [Note : For reaction (a) even if the candidate writes ⁷₃X, award 1 mark]

278.

Complete the following nuclear reactions :(a)10 5B + 01n → 24He + ........(b) 9442Mo + 21n → 9543Te + .........

Answer»

(a)¹⁰₅B + ₀¹n → ₂⁴He + ⁷₃Li

(b) ⁹⁴₄₂Mo + ²₁n → ⁹⁵₄₃Te + ¹₀n

[Note : For reaction (a) even if the candidate writes

⁷₃X, award 1 mark]

Discussion

279.	(23592U + 10n → 14456Ba + 8936kr + 10n +energy)(Give full credit for any other one correct example.)
Answer» The number are conserved but the total mass is nor conserved. The total mass of the free protons/neutrons is more than their total mass within the nucleus. The lost mass (= Δm), gets converted into energy as per the relation. E = (Δ m)c² [Also, accept alternative ways of explaining the phenomenon]

279.

(23592U + 10n → 14456Ba + 8936kr + 10n +energy)(Give full credit for any other one correct example.)

Answer»

The number are conserved but the total mass is nor conserved.

The total mass of the free protons/neutrons is more than their total mass within the nucleus. The lost mass (= Δm), gets converted into energy as per the relation. E = (Δ m)c²

[Also, accept alternative ways of explaining the phenomenon]

Discussion

280.	In the reactions given below :(i) 116C → zyB + x + v(ii) 126C + 126C → 20aNe + cbHeFind the values of x, y, and z and a, b and c.
Answer» (i) x = β⁺ ⁰₁e, y = 5, z = 11 (ii) a = 10, b = 2c =14

280.

In the reactions given below :(i) 116C → zyB + x + v(ii) 126C + 126C → 20aNe + cbHeFind the values of x, y, and z and a, b and c.

Answer»

(i) x = β⁺ ⁰₁e, y = 5, z = 11

(ii) a = 10, b = 2c =14

Discussion

281.	Pick out the correct statements from the following I. Electron emission during `Brta`-decay is always accompanied by neutrons. II. Nuclear force is charge independent. III. Fusion is the chief source of stellar energy.A. Both I and II are correctB. Both I and III are correctC. Only I is correctD. Both II and III are correct
Answer» Correct Answer - D We know that nuclear forces are charge independent and the chief sourec of stellar energy is fussion.

Discussion

282.	Obtain the maximum kinetic energy of `beta`- particles, and the radiation frequencies of `gamma` decays in the decay scheme shown in Fig. You are given that `m(.^(198)Au)=197.968233 u` `m(.^(198)Hg)=197.966760 u`.
Answer» Energy corresponding to `r_(1)` `E_(1)=1.088-0=1.088 MeV` `= 1.088xx1.6xx10^(-13)` Joule Frequency `v_(1)=(E_(1))/(h)` `= (1.088xx1.6xx10^(-13))/(6.6xx10^(-34))` `= 2.63xx10^(20)Hz` similary `v_(2)=(E_(2))/(h)` `= (0.412xx1.6xx10^(-12))/(6.6xx10^(10))` `= 9.98xx10^(13)Hz` and `v_(3)=(E_(3))/(h)` `= ((1.088-0.412)xx1.6xx10^(-13))/(6.6xx10^(20)Hz)` Maximum K.E. of `beta_(1)` particle `K_("max")(beta_(1))=[m(._(79)Au^(198))-` mass of second excited state of `._(80)Hg^(198)]xx931 MeV` `= [m(._(79)Au^(198))-m(._(82)Hg^(198))-(1.088)/(931)]xx931 MeV` `= 931[197.968233-197.966760]-1.088 MeV` `= 1.371-1.088=0.283 MeV` similarly `k_("max")(beta_(2))=0.957 MeV`

Discussion

283.	Density of all the nuclei is same. Radius of nucleus is directly proportional to the cube root of mass number.A. If both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason are true but Reason is not correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true.
Answer» Correct Answer - A

Discussion

284.	Assertion:When a nucleus is in an excited state, it can make a transition to a lower energy state by the emission of gamma rays . Reason:These are energy levels for a nucleus just like there are energy levels in atoms .A. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false
Answer» Correct Answer - b

Discussion

285.	The relationship between decay constant `lamda` and half-life T of a radioactive substance isA. `lamda=(log_(10)2)/(T)`B. `lamda=(log_(e)2)/(T)`C. `lamda=(log_(2)10)/(T)`D. `lamda=(log_(2)e)/(T)`
Answer» Correct Answer - B The relation between decay constant `lamda` and half-life T can be given as `T=(0.693)/(lamda)` `:.T=(log_(e))/(lamda)`

Discussion

286.	For the stabilty of any nucleus,A. binding energy per nucleon will be moreB. binding energy per nucleon will be lessC. number of electrons will be moreD. None of the aobve
Answer» Correct Answer - A The binding energy of a nucleus is the energy required to take its nucleus away from one another. It is generally expressed as binding energy per nucleus.Higher the binding energy per nucleon, more stable is the nucleus.

Discussion

287.	The binding energy per nucleon of `._(3)^(7) Li` and `._(2)^(4)He` nuclei are `5.60` MeV and `7.06` MeV, respectively. In the nuclear reaction `._(3)^(7)Li+._(1)^(1)H rarr ._(2)^(4)He+._(2)^(4)He+Q`, the value of energy `Q` released isA. `19.6` Me VB. `-2.4` Me VC. `8.4` Me VD. `17.3` Me V
Answer» Correct Answer - D The binding energy for `""_(1)H^(1)` is around zero and also not given in the question so we can igonor it `Q=2(4xx7.06)-7xx(5.60)=(8xx7.06)-(7xx5.60)` `=56.48-39.2=17.28MeV~=17.3 MeV`

Discussion

288.	A radioactive substance emits n beat particles in the first 2 s and `0.5` n beta particles in the next 2 s. The mean life of the sample isA. 4 sB. 2 sC. `(2)/((1n 2))`D. 2 (1n 2) s
Answer» Correct Answer - C We have, `lamda=(0.693)/(T_(1//2)` Mean life, `tau=(1)/(lamda)=(t_(1//2))/(0.693)=(2)/(2.303log_(2))=(2)/(In2)s`

Discussion

289.	In the depletion region of a diode.A. there are no mobile chargesB. equal number of holes and elections exist, making the regin neutalC. recombination of holes and electrons has taken placeD. immobile charged ions exist.
Answer» Correct Answer - A::B::D The space-charge regions on both the sides of `p - n` junction which has immobile ions and entirely lacking of any charge carries will form a region called depletion region of a diode. The number of ionized acceptors on the `p-`side equal the number of ionized donors on the `n-`side

Discussion

290.	The amplifiers `X, Y` and `Z` are connected in series. If the voltage gains of `X, Y` and `Z` are `10, 20` and `30`, respectively and the input signal is `1mV` peak value, then what is the output signal voltage (peak value) (i) if dc supply voltage is 10V ? (ii) if dc supply voltage is 5V?
Answer» given, `AV_(x) = 10, AV_(y) = 20, AV_(z) = 30`, `Delta V_(1) = 1 mV = 10^(-3) V` Now, `("Output Signal Voltage " (Delta V_(0)))/("Input Singal Voltage " (Delta V_(i))) =` Total voltage amplification `= Av_(x) xx Av_(y) xx Av_(z)` `rArr Delta V_(0) = Av_(x) xx Av_(y) xx Av_(z) xx Delta V_(i)` `10 xx 20 xx 30 xx 10^(-3) = 6V` (i) If `DC` supply voltage is 10 V, then output is 6V, since theoretical gain is equal to practical gain, i.e., ouput can never be greater than 6V. (ii) If DC supply voltage is 5V, i.e., `V_(c c) = 5V`. Then, output peak will not exceed 5V. Hence `V_(0) = 5V`.

Discussion

291.	(i) Name the type of a diode whose characteristics are shown in figure (a) and (b) (ii) What does the point P in fig. (a) represent ? (iii) What does the point P and Q in fig. (b) represent ?
Answer» The characteristic curve (a) is of Zener diode and curve (b) is of solar cell. (ii) The point p in fig. (a) represents Zener break down voltage. (iii) In fig. (b), the point Q represents zero voltage and negative current, it means light falling on solar cell with ateast minimum threshold frequency gives the current in opposite direction to the due to a battery connected to solar cell. But for the point Q, the batter is short circuited. Hence represents the short circuit current. In fig. (b), the point P represents some positive voltage on solar cell with zero current through solar cell. It means, there is a battery connected to a solar cell which gives rise to the equal an opposite current to that in solar cell by virtue of light falling on it. As current is zero for point P, hence we say P respresents open circuit voltage.

Discussion

292.	In a CE transistor amplifier, there is a current and voltage gain associated with the circuit. In other words there is a power gain. Considering power a measure of energy, does the circuit violate conservation of energy ?
Answer» In CE transistor amplifier, the power gain is very high. In this circuit, the extra power required for amplified output is obtained from DC source. Thus, the circuit used does not violet the law of conservation.

Discussion

293.	Complete the given nuclear reaction
Answer» _zX^A + -1^e → _z-1X^A + v + Energy

Discussion

294.	A beam of radio active radiation is unaffected in a combined electric and magnetic field in mutually perpendicular direction. A boy argues that, it is essentially a gamma ray. Do you agree with him. Justify your answer.
Answer» It need not be gama ray. It can be α or β ray as the electric and magnetic field are in mutually perpendicular direction.

Discussion

295.	Suppose you are a health physicist and you are being consulted about a spill occurred in a radio chemistry lab. The isotope spilled out in the lab was 500 micro cure of 131Ba which has half-life of 12 – days.1. What is the decay constant of 131Ba2. Your recommendation is to clear the lab until the radiation level is down to 1 micro curie. How length will the lab have to be closed.
Answer» 1. λ = \(\frac{0.693}{T_{1/2}} =\frac{0.693}{12\times24\times60\times60}\) = 6.6 x 10^-7 2. 9 × 12 = 108 days.

295.

Suppose you are a health physicist and you are being consulted about a spill occurred in a radio chemistry lab. The isotope spilled out in the lab was 500 micro cure of 131Ba which has half-life of 12 – days.1. What is the decay constant of 131Ba2. Your recommendation is to clear the lab until the radiation level is down to 1 micro curie. How length will the lab have to be closed.

Answer»

1. λ = \(\frac{0.693}{T_{1/2}} =\frac{0.693}{12\times24\times60\times60}\) = 6.6 x 10^-7

2. 9 × 12 = 108 days.

Discussion

296.	Figure below represents radiation coming out from a radioactive element.1. Identify the radiation B. Give reason.2. If the electric field is replaced by magnetic field perpendicular and into the plane of the paper, identify the particle deflecting towards right.
Answer» 1. Gamma-ray. It has no change 2. C which is β ray

296.

Figure below represents radiation coming out from a radioactive element.1. Identify the radiation B. Give reason.2. If the electric field is replaced by magnetic field perpendicular and into the plane of the paper, identify the particle deflecting towards right.

Answer»

1. Gamma-ray. It has no change

2. C which is β ray

Discussion

297.

Match the following.ABi. Isotopes17Cl37 , 19KI39ii. Isobars8O16,8O1iii. Isotones1H3,2He3iv. 1H3,3Lil

Answer»

A	B
i. Isotopes	₈O¹⁶,₈O¹
ii. Isobars	₁H³,₂He³
iii. Isotones	₁₇Cl³⁷, ₁₉KI³⁹

Discussion

298.	A common example of `beta-"decay"` is `""_(15)P^(32)to""_(16)P^(32)+x+y` Then, x and y stand forA. electron and nectrinoB. positron and neutrinoC. electron and antineutrinoD. positron and antineutrino
Answer» Correct Answer - C In `beta^(-)` deacy a neutron in the nucleus is transformed into a proton, an electron and an anti-neutrino `nto P + bar(e)+bar(v)` So, x and y, stand for electron and anti-neutrino.

Discussion

299.	From the relation R = R0 A1/3 , Where R0 is a constant and A is the mass number of a nucleus show that the nuclear matter density is nearly constant (i.e. Independent of A).
Answer» Density of nucleus(p) It is defined as the nuclear mass per unit volume. i.e ρ = \(\frac{mass\,of\,one\,nucleus}{volume\,of\,nucleus}\) = \(\frac{mass\,of\,one\,nucleon\times number\,of\,nucleons}{\frac{4}{3}\pi r^3}\) = \(\frac{1.66\times10^{-27}A}{\frac{4}{3}\pi (R_0A^{1/3})^3}\) = \(\frac{3\times1.66\times10^{-27}A}{4\times3.142\times(1.2\times10^{-15})^3\times A}\) = 2.29 x 10¹⁷Kgm^-3 Thus the nuclear density is of the order of 10¹⁷kgm^-3 and is independent of its mass number. Therefore, all nuclei have the same approximate density.

299.

From the relation R = R0 A1/3 , Where R0 is a constant and A is the mass number of a nucleus show that the nuclear matter density is nearly constant (i.e. Independent of A).

Answer»

Density of nucleus(p) It is defined as the nuclear mass per unit volume.

i.e ρ = \(\frac{mass\,of\,one\,nucleus}{volume\,of\,nucleus}\)

= \(\frac{mass\,of\,one\,nucleon\times number\,of\,nucleons}{\frac{4}{3}\pi r^3}\)

= \(\frac{1.66\times10^{-27}A}{\frac{4}{3}\pi (R_0A^{1/3})^3}\) = \(\frac{3\times1.66\times10^{-27}A}{4\times3.142\times(1.2\times10^{-15})^3\times A}\)

= 2.29 x 10¹⁷Kgm^-3

Thus the nuclear density is of the order of 10¹⁷kgm^-3 and is independent of its mass number. Therefore, all nuclei have the same approximate density.

Discussion

300.	Give the relation between half-life and mean life
Answer» t_1/2 = 0.693τ.

Discussion

Explore topic-wise InterviewSolutions in .

Beta rays emitted by a radicactive material areA. Electromagnetic radiationB. Neutral particlesC. The electrons orbiting around the nucleusD. Charged particles emitted by nucleus

When a deuterium is bombarded on `._(8)O^(16)` nucleus, an `alpha`-particle is emitted, then the product nucleus isA. `._7N^13`B. `._5B^10`C. `._4Be^9`D. `._7N^14`

Boron rods are used in nuclear reactors becauseA. boron can absorb neutronsB. strength is given to the plantC. chain reaction of neutrons and `""_(92)U^(240)`D. the reactors look good

The half life of `._92U^(238)` against `alpha` -decay is `4.5xx10^9` years. What is the activity of 1g sample of `._92U^(238)`?

After 300 days, the activity of a radioactive sample is 5000 dps (disintegrations per sec). The activity becomes 2500 dps after another 150 days. The initial activity of the sample in dps isA. 20000B. 10000C. 7000D. 25000

In the given reaction `._z X^A rarr ._(z+1)Y^A rarr ._(z-1) K^(A - 4) rarr ._(z-1) K^(A - 4)` Radioactive radiations are emitted in the sequence.A. `alpha,betagamma`B. `gamma,alpha,beta`C. `beta,alpha,gamma`D. `lamda,beta,alpha`

In the given reaction `._z X^A rarr ._(z+1)Y^A rarr ._(z-1) K^(A - 4) rarr ._(z-1) K^(A - 4)` Radioactive radiations are emitted in the sequence.A. `alpha,beta,gamma`B. `gamma,alpha,beta`C. `beta,alpha,gamma`D. `beta,gamma,alpha`

A radioactive element X with half life 2 h decays giving a stable element Y. After a time t, ratio of X and Y atoms is 1:16 .Time t isA. 6 hB. 4 hC. 8 hD. 16 h

What fissionable material was used in the bomb dropped at Nagasaki (Japan) in the year 1945 ?A. UraniumB. NeptuniumC. BerkeliumD. Plutonium

Write the truth table for the circuit shown in figure given below. Name the gate that the circuit resembles.

During a negative beta decay,A. an atomic electron is ejectedB. an electron which is already present with in the nucleus is ejectedC. a neutron in the nucleus decays emitting on electronD. a part of the binding energy is converted into electron

A radioactive nucleus `._92 X^235` decays to `._91 Y^231`. Which of following particles are emitted ?A. One alpha and one electronB. Two deuterons and one positronC. One alpha and one protonD. One proton and four neutrons

Calculate the binding energy and binding energy per nucleon in MeV for carbon-12 nucleus. Given that mass of the proton is 1.00727 amu while the mass of the neutron is 1.00866 amu.

What is the binding energy per nucleon of `_(6)C^(12)`nucleus? Given , mass of `C^(12)` `(m_(c))_(m)` = 12.000 u Mass of proton `m_(p)` = 1.0078 u Mass of neutron `m_(n)` = 1.0087 u and 1 amu = 931.4 MeVA. 5.26 MeVB. 6.2 MeVC. 4.65 MeVD. 7.68 MeV

Define Becquerel and Curie.

What is the role of controlling rods in a nuclear reactor ?

Why are nuclear fusion reactions called thermo nuclear reactions ?

Which of the following equactions pick out the possible nuclear fusion reactions?A. `._6^13C +._1^1H to ._6^14C` +4.3 MeVB. `._6^12C + ._1^1H to ._7^13N` + 2 MeVC. `._7^14N + ._1^1H to ._8^15O ` + 7.3 MeVD. `._92^235C + ._0^1N to ._54^140 Xe + ._38^94Sr + ._0^1n + ._0^1n` + 200 MeV

`A U^(235)` reactor generated power at a rate of P producting `2xx10^(18)` fussion per second. The energy released per fission is 185 MeV. The value of P isA. `59.2` MWB. `370xx10^(18)MW`C. `0.59 MW`D. 370 MW

If `200 MeV` energy is released in the fission of a single `U^235` nucleus, the number of fissions required per second to produce `1` kilowatt power shall be (Given `1 eV = 1.6 xx 10^-19 J`).A. `3.125xx10^13`B. `1.52xx10^6`C. `3.125xx10^12`D. `3.125xx10^14`

In a nuclear reactor, the number of `U^(235)` nuclei undergoing fissions per second is `4xx10^(20).` If the energy releases per fission is 250 MeV, then the total energy released in 10 h is `(1 eV= 1.6xx10^(-19)J)`A. `576xx10^(6)J`B. `576xx10^(12)J`C. `576xx10^(15)J`D. `576xx10^(12)J`

Complete the following nuclear reactions :(i) 20884Po → 20482Pb + ........(ii) 3215P → 3216S + ..........

Complete the following nuclear reactions :(a)10 5B + 01n → 24He + ........(b) 9442Mo + 21n → 9543Te + .........

(23592U + 10n → 14456Ba + 8936kr + 10n +energy)(Give full credit for any other one correct example.)

In the reactions given below :(i) 116C → zyB + x + v(ii) 126C + 126C → 20aNe + cbHeFind the values of x, y, and z and a, b and c.

Obtain the maximum kinetic energy of `beta`- particles, and the radiation frequencies of `gamma` decays in the decay scheme shown in Fig. You are given that `m(.^(198)Au)=197.968233 u` `m(.^(198)Hg)=197.966760 u`.

The relationship between decay constant `lamda` and half-life T of a radioactive substance isA. `lamda=(log_(10)2)/(T)`B. `lamda=(log_(e)2)/(T)`C. `lamda=(log_(2)10)/(T)`D. `lamda=(log_(2)e)/(T)`

For the stabilty of any nucleus,A. binding energy per nucleon will be moreB. binding energy per nucleon will be lessC. number of electrons will be moreD. None of the aobve

The binding energy per nucleon of `._(3)^(7) Li` and `._(2)^(4)He` nuclei are `5.60` MeV and `7.06` MeV, respectively. In the nuclear reaction `._(3)^(7)Li+._(1)^(1)H rarr ._(2)^(4)He+._(2)^(4)He+Q`, the value of energy `Q` released isA. `19.6` Me VB. `-2.4` Me VC. `8.4` Me VD. `17.3` Me V

A radioactive substance emits n beat particles in the first 2 s and `0.5` n beta particles in the next 2 s. The mean life of the sample isA. 4 sB. 2 sC. `(2)/((1n 2))`D. 2 (1n 2) s

In the depletion region of a diode.A. there are no mobile chargesB. equal number of holes and elections exist, making the regin neutalC. recombination of holes and electrons has taken placeD. immobile charged ions exist.

The amplifiers `X, Y` and `Z` are connected in series. If the voltage gains of `X, Y` and `Z` are `10, 20` and `30`, respectively and the input signal is `1mV` peak value, then what is the output signal voltage (peak value) (i) if dc supply voltage is 10V ? (ii) if dc supply voltage is 5V?

(i) Name the type of a diode whose characteristics are shown in figure (a) and (b) (ii) What does the point P in fig. (a) represent ? (iii) What does the point P and Q in fig. (b) represent ?

In a CE transistor amplifier, there is a current and voltage gain associated with the circuit. In other words there is a power gain. Considering power a measure of energy, does the circuit violate conservation of energy ?

Complete the given nuclear reaction

A beam of radio active radiation is unaffected in a combined electric and magnetic field in mutually perpendicular direction. A boy argues that, it is essentially a gamma ray. Do you agree with him. Justify your answer.

Figure below represents radiation coming out from a radioactive element.1. Identify the radiation B. Give reason.2. If the electric field is replaced by magnetic field perpendicular and into the plane of the paper, identify the particle deflecting towards right.

Match the following.ABi. Isotopes17Cl37 , 19KI39ii. Isobars8O16,8O1iii. Isotones1H3,2He3iv. 1H3,3Lil

A common example of `beta-"decay"` is `""_(15)P^(32)to""_(16)P^(32)+x+y` Then, x and y stand forA. electron and nectrinoB. positron and neutrinoC. electron and antineutrinoD. positron and antineutrino

From the relation R = R0 A1/3 , Where R0 is a constant and A is the mass number of a nucleus show that the nuclear matter density is nearly constant (i.e. Independent of A).

Give the relation between half-life and mean life