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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
When `p- n` junction diode is forward baised themA. both the deplection regain and harrier height are reductedB. the depletion regain is widened and harrier height is reducedC. the deplection regain , is reducted and harrier heighis increasesD. Both the depletion regain and barrier are increases |
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Answer» Correct Answer - A Both the deletion regain and harrier beight is reduced |
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| 102. |
If radius of the `_(13)^(27) Al` necleus is estimated to be `3.6` fermithen the radius of `_(52)^(125)Te` nucleus be nearlyA. `8 fermi`B. `6 fermi`C. `5 fermi`D. `6 fermi` |
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Answer» Correct Answer - B KEY CONCEPT : `R = R_(0)(A)^(1//3)` `:. (R_(1))/(R_(2)) = (A_(1))/(A_(2))^(1//3) = ((27)/(125))^(1//3) = (3)/(5)` `R_(2) = (5)/(3) xx 3.6 = 6 fermi` |
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| 103. |
The ratio of de - Broglie wavelength of `alpha `- particle to that of a proton being subjected to the same magnetic field so that the radii of their path are equal to each other assuming the field induction vector `vec(B)` is perpendicular to the velocity vectors of the `alpha` - particle and the proton isA. `1`B. `1//4`C. `1//2`D. `2` |
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Answer» When a charged particle of charge `q`, mass `m` enters perpendicularly to the magnetic induction `vecB` of a magnetic field, it will experience a magnetic force `F=q(vecvxxvecB)=q vB sin 90^(@)=qvB` that provide a centripetal acceleration `(v^(2))/(r )` `rArr qvB=(mv^(2))/(r )rArrmv=qBr` `rArr` The `de`-Broglie wavelength `lambda=(h)/(mv)=(h)/(qBr)` `rArr (lambda_(alpha-"particle"))/(lambda_("proton"))=(q_(p)r_(p))/(q_(alpha)r_(alpha))` Since `(r_(alpha))/(r_(p))=1` and `(q_(alpha))/(q_(p))=2` `rArr (lambda_(alpha))/(lambda_(p))=1//2` |
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| 104. |
The edge of a cube is `a=1.2xx10^(-2) m`. Then its volume will be recorded as:A. `1.728xx10^(-6)m^(3)`B. `1.72xx10^(-6)m^(3)`C. `1.7xx10^(-6)m^(3)`D. `0.72xx10^(-6)m^(3)` |
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Answer» `v=l^(3)=1.728xx10^(-6)` Length has two significant figure `v=1.7xx10^(-6)m^(3)` |
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| 105. |
A physical quantity `x` is calculated from the relation `x = ( a^(2) b^(3))/(c sqrt( d))`. If the percentage error in `a, b , c , and d are 2% , 1% , 3%, and 4%`, respectively , what is the percentage error in `x`?A. `+-11%`B. `+-13%`C. `+-12%`D. `+-14%` |
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Answer» `(Deltax)/(x)= +-((2Deltaa)/(a)+(3Deltab)/(b)+(Deltac)/(c )+(Deltad)/(2d))` `=+-(2xx2+3xx1+3+(1)/(2)xx4)= +- 12%` |
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| 106. |
A moving hydrogen atom makes a head on collision with a stationary hydrogen atom. Before collision both atoms are in in ground state and after collision they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excited state? |
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Answer» Correct Answer - 2 |
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| 107. |
A moving hydrogen atom makes a head on collision with a stationary hydrogen atom. Before collision both atoms are in in ground state and after collision they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excited state?A. `20.4eV`B. `10.2eV`C. `54.4eV`D. `13.6eV` |
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Answer» Correct Answer - A `K = (P^2) /(2m)` in collision, momentum p remains constant. `:. K prop (1)/(mass)` After collision, mass has doubled. So kinetic energy will remain `(K)/(2).` Hence,loss is also `(K)/(2)` Now, `(K)/(2)` = minimum excitation energy required. =10.2 eV `rArr K = 20.4 eV` |
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| 108. |
The difference in the variation of resistence with temperature in a metel and a semiconductor arises essmially due to the difference in theA. crystal sturctureB. variation of the number of change carriers with tempeatureC. type of bondingD. variation of scattaring mechanism with temperature |
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Answer» Correct Answer - B When the temperature increases , certain bounding electrons become free which tend to promate conductivity simultaneously number of collsion between electron and positive kernels increase |
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| 109. |
A radioactive sample at any instant has its disintegration ratye `5000` disintegrations per minute After `5` minutes , the rate is `1250` disintegration per Then , the decay constant (per minute)A. `0.4 in 2`B. `0.2 in 2`C. `0.1 in 2`D. `0.8 in 2` |
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Answer» Correct Answer - A `lambda = (1)/(t) log_(e) (A_(0))/(A) = (1)/(5) log _(e) (5000)/(1250) = 0.4 log _(e) 2 ` |
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| 110. |
A parallel beam of monochromatic light of wavelength 663 nm is incident on a totally reflection plane mirror. The angle of incidence is `60^@` and the number of photons striking the mirror per second is `1.0xx10^19`. Calculate the force exerted by the light beam on the mirror. |
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Answer» Correct Answer - A Force = Rate of change of momentum `=2 [N] [(h)/(lambda)]. cos 60^@` N = number of photons striking per second `(h)/(lambda) = momentum of one photn. |
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| 111. |
A source emits monochromatic light of frequency `5.5xx10^(14)` Hzat a rate of 0.1 W. Of the photons given out, 0.15% fall on the cathode of a photocell which gives a current of `6muA` in an extrnal circuit. (a) Find the enrgy of a photon. (b) Find the number of photons leaving the source per second. (C) Find the percentage of the photons falling on the cathode which produce photoelectrons. |
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Answer» Correct Answer - A::B::C (a)`E = hf= (6.6xx10^(-034))(5.5xx10^(14))` `=36.3xx10^(-20) J` =2.75 eV (b) Number of photons leaving the source per second, n = (p)/(E ) = (0.1)/(36.3xx10^(-20) `=2.75xx10^(-17)` (c ) Number of photons falling on cathode per second `n_1 = (0.15)/(100)xx2.75xx10^(17)` `=4.125x10^(-14)` Number of photonelectrons emitting per second, `n_2 = (6xx10^(-6))/(1.6xx10^(-19)) = 3.75xx10^(13)` `:. % = (n_2)/(n_1)xx100 = (3.75)/(4.125x10^(14)xx100` =9%. |
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| 112. |
An electron is confined to a tube of length L. The electron’s potential energy in one half of the tube is zero, while the potential energy in the other half is 10eV. If the electron has a total energy `E = 15 eV`, then the ratio of the deBroglie wavelength of the electron in the 10eV region of the tube to that in the other half is -A. `1//sqrt3`B. `sqrt3`C. 3D. `(1)/(3)` |
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Answer» Correct Answer - B |
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| 113. |
A monochromatic beam of light `(lamda=4900A)` incident normally upon a surface produces a pressure of `5xx10^(-7)Nm^(-2)` on it. Assume that `25%` of the light incident in reflected and the rest absorbed. Find the number of photons falling per second on a unit area of thin surface |
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Answer» `P=[2(0.25)+0.75](I)/(c )=1.25(I)/(c )` `:. ` Intensity of light, `I=(cP)/(1.25)=((3xx10^(8))(5xx10^(-7)))/(1.25)=120Wm^(-2)` Energy of photon, `E=(hc)/(lambda)=((6.63xx10^(-34))(3xx10^(8)))/(0.49xx10^(-6))=4xx10^(-19)J` `:. ` Number of photons incident per unit area per second `n=(I)/(E)=(120)/(4xx10^(-19))=3xx10^(20)m^(-2)s^(-1)` |
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| 114. |
In previous question KE of daughter nucleous is -A. 3.16 MevB. 4.16 MevC. 5.16 MevD. 6.16 Mev |
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Answer» Correct Answer - D |
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| 115. |
In a hypothetical atom, potential energy between electron and proton at distance r is given by `((-ke^(2))/(4r^(2)))` where k is a constant Suppose Bohr theory of atomic structrures is valid and n is principle quantum number, then total energy E is proportional toA. `n^(5)`B. `n^(2)`C. `n^(6)`D. `n^(4)` |
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Answer» Correct Answer - C |
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| 116. |
The kinetic energy of an electron is E when the incident wavelength is `lamda` To increase ti KE of the electron to 2E, the incident wavelength must beA. `2lambda`B. `lambda//2`C. `((hclambda))/((Elambda+hc))`D. `((hclambda))/((Elambda-hc))` |
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Answer» Correct Answer - C |
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| 117. |
The wavelengths of `K_alpha` X-rays from lead isotopes `Pb^(204) , Pb^(206) and Pb^(208)` are `lambda_1, lambda_2 and lambda_3` respectively. Choose the correct alternative.A. `lambda_(1)=lambda_(2)=lambda_(3)`B. `lambda_(1)gtlambda_(2)gtlambda_(3)`C. `lambda_(1)ltlambda_(2)ltlambda_(3)`D. `lambda_(2)=sqrt(lambda_(1)lambda_(3))` |
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Answer» Correct Answer - A::D |
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| 118. |
Which of the following is a correct statement?A. (a) Beta rays are same as cathode raysB. (b) Gamma rays are high energy neutronsC. (c) Alpha particles are singly ionized helium atomsD. (d) Protons and neutrons have exactly the same mass |
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Answer» Correct Answer - A Both the beta rays and the cathode rays are made up of electrons. So, only option (a) is correct. (b) Gamma rays are electromagenetic waves. (c) Alpha particles are doubly ionized helium atoms and (d) Protons and neutrons have approximately the same mass. Therefore, (b), (c) and (d) are wrong options. |
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| 119. |
The electron emitted in beta radiation originates fromA. (a) inner orbits of atomB. (b) free electrons existing in nucleiC. (c) decay of a neutron in a nucleusD. (d) photon escaping from the nucleus |
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Answer» Correct Answer - C During `beta`-decay, a neutron is transformed into a proton and an electron. This is why atomic number (Z=number of protons) increases by one and mass number (A=number of protons + neutrons) remains unchanged during `beta`-decay. |
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| 120. |
During a negative beta decay,A. (a) an atomic electron is ejectedB. (b) an electron which is already present within the nucleus is ejectedC. (c) a neutron in the nucleus decays emitting an electronD. (d) a part of the binding energy of the nucleus is converted into a electron |
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Answer» Correct Answer - C Following nuclear reaction takes place `_0n^1-rarr_1H^1+_-1e^0+barv` `barv` is antineutrino. |
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| 121. |
Dividing a negative beta dacayA. as atomic electron is ejectionB. as electron which is already present within the nuclease is ejectionC. a neclues in the nuclease decay emiting an electronD. a part of the necule the binding energy of the nuclease is converted into an electron |
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Answer» Correct Answer - C in the correct option `lambda_(min)= (hc)/(eV)` |
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| 122. |
The largest wavelength in the ultraviolet region of the hydrogen spetrum is `122` nm. The samallest wavelength in the infrared ragaion of the hydrogen spetrum (to the nearest integer) isA. `802 nm`B. `823 nm`C. `1882 nm`D. `1648 nm` |
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Answer» Correct Answer - B The skallest , frequency end large wavwlength in ullraviolet region will be for transtion of from orbit `2` to orbit`1` `:. (1)/(lambda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` `rArr (1)/(122 xx 10^(-9)m) = R[(1)/(1^(2)) - (1)/(2^(2))] = R [(1 - (1)/(4)] = (3R)/(4)` E = (4)/(3 xx 122 xx 10^&(-9)) m^(-1)` The highest frequency and smaller wavelength for infrared reguion will be for transtionfrom `oo` ti 3rd orbit `:. (1)/(lambda) = R((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` `rArr (1)/(lambda) = (4)/(3 xx 122 xx 10^(-9)) ((1)/(3^(2)) - (1)/(oo))` `:. lambda = (3 xx 122 xx 9 xx 10^(-9))/(4) = 823.5 nm` |
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| 123. |
During a nuclear fasion reactionA. a heavy nuclease break into two fragments by itselfB. a light nucleus bomberded by theremal neutrous breaks upC. a heavy nucleus bombered by thermal neudrons break upD. two light nucleus combered by thermal and posibly other product |
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Answer» Correct Answer - D in the correct option `lambda_(min)= (hc)/(eV)` |
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| 124. |
It is propssed to use the nucles fasion `_(1)^(2) H +_(1)^(2) H rarr _(2)^(4) He` in a nucleas of `200MW` rating if the energy from the above reaction is used with a `25` per cast effecincy in the rector , low maney game of deuterium fiel will be needed per day (The masses of `_(1)^(2)H and _(4)^(2) He are 2.0141`atomic mass unit and `4.0028`atyomic mass uniot repertively) |
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Answer» Correct Answer - A energy required per day ` E = p xx 1 = 200 xx 10^(9) xx 24 xx 60 xx 60` `= 1.728 xx 10^(13)J` energy released per fusion reaction `= [ 2 (2.0141)-4.0026] xx 931.5 MeV` `= 23.15 xx 10^(-13)J` `:. ` No of fusion reactions required `= (1.728 xx 10^(13))/(38.15 xx 10^(-13)) = 0.045 xx 10^(26)` :.` No of deuterium atoms required `= 2 xx 0.045 xx 10^(26) = 0.09 xx 10^(26)` number of deterium atom `= (0.09 xx 10^(26))/(6.02 xx 10^(23)) = 14.95` `:.`Mass in gram of deterium atom `= 14.95 xx 2 = 29.9 g` But the efficiency is `25%` Therefore, the efficiency mass required` = 199.6 g` |
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| 125. |
In the option given below , let `E` denote the rest mass energy of a nucleas and `n` a neutron .The correct option isA. `E(`_(92)^(236)U) gt`E(`_(92)^(236)I) +`E(`_(92)^(236)Y) + 2E (n)`B. `E(`_(92)^(236)U) lt`E(`_(92)^(137)I) +`E(`_(39)^(97)Y) + 2E (n)`C. `E(`_(92)^(236)U) lt`E(`_(56)^(140)Ba) +`E(`_(36)^(94)Kr) + 2E (n)`D. `E(`_(92)^(236)U) =`E(`_(56)^(140)Ba) +`E(`_(94)^(36)Kr) + 2E (n)` |
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Answer» Correct Answer - A Ioline and `Y`thrium` are medium sized nuclius and therefore have more binding energy per nuclius as compared to Uranium which has a big nuclus and less `B.E. `incleon. In after work , lodine and Yurinum are more and therefore posses less energy end rest Also when Uranium nucles having kinetic energies. |
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| 126. |
The rest mass of a deuteron is equivalent to an energy of `1876 MeV`, that of a proton to `939 MeV`, and that of a neutron to` 940 MeV. A deutron may disintegrate to a proton and neutron ifA. emits a y-ray photon of energy 2 MeVB. captures a y-ray photon of energy 2MeVC. emits a y-ray photon of energy o3 MeVD. captures a y-ray photon energy 3 MeV |
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Answer» Correct Answer - D |
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| 127. |
in an experimental set up to study the photoelectric effect a point soure fo light of power `3.2xx10^(-3)` W was taken. The source can emit monoenergetic photons of energy 5eV and is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work-function 3.0 eV. The radius of the sphere is `r = 8..10^(-3)` m. The efficiency of photoelectric emission is one for every `10^6` incident photons. Based on the information given above answer the questions given below. (Assume that the sphere is isolated and photoelectrons are instantly swepts away after the emission). de-Broglie wavelength of the fastest moving photoelectron isA. `6.63 Å`B. `8.69 Å`C. `2 Å`D. `5.26 Å` |
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Answer» Correct Answer - B `K_(max) = E -W = 2eV `lambda = (sqrt(150)/(KE(in eV)), for an electron `=sqrt(150)/(2) ~~8.6 Å` |
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| 128. |
in an experimental set up to study the photoelectric effect a point soure fo light of power `3.2xx10^(-3)` W was taken. The source can emit monoenergetic photons of energy 5eV and is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work-function 3.0 eV. The radius of the sphere is `r = 8..10^(-3)` m. The efficiency of photoelectric emission is one for every `10^6` incident photons. Based on the information given above answer the questions given below. (Assume that the sphere is isolated and photoelectrons are instantly swepts away after the emission). It was observed that after some time emission of photoelectrons from the sphere stopped. Charge on the sphere when the photon emission stops isA. `16piepsilon_0^r` coulombB. `8piepsilon_0^r` coulombC. `15piepsilon_0^r` coulombD. `20piepsilon_0^r` coulomb |
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Answer» Correct Answer - B `K_(max)` is 2 eV. Hence, stopping potential is 2V. Photoemission stops when potential of sphere becomes 2V. `:. 2 =(q)/(4piepsilon_0r)` `:. q = 8piepsilon_0 r.` |
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| 129. |
The wavelength for n=3 to n=2 transition of the hydrogen atom is 656.3 nm. What are the wavelength for this same transition in (a) positronium, which consists of an electron and a positron (b) singly ionized helium |
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Answer» Correct Answer - A::B::C::D (a) Reduced mass of positronium and electron is `(m)/(2),` where, m = mass of electron `E prop m :. lambda prop (1)/(m)` m has become half, so `lambda` will become two times or 1312 nm or `1.31 mum.` . For singly ionixed helium atom Z =2 `:. lambda is (1)/(4) th or 164 nm` |
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| 130. |
The time taken by a photoelectron to come out after the photon strikes is approximatelyA. `10^(-4)s`B. `10^(-10)s`C. `10^(-16)s`D. `10^(-1)s` |
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Answer» Correct Answer - B The order of time is `10^(-10)5` |
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| 131. |
When a particle is restricted to move along x- axis between `x = 0 and x = 4 ` whwre a is opf nanometer demension , its energy can take only certain spscfic values . The allowed energies of the particles only in such a restiricted regain , correspond to the formation of standing wave with nodes at its end ` x = 0 and x = a `.The wavelength of this standing wave is related to the linear momentum p of the paarticle according to the de Broglie relation .The energy of the particle of mass `m` is reated to its linear momentum as `E = (p^(2))/(2m)` . thus , the energy of the particle can be denoted by a quantum number `n` taking value `1,2,3,....(n= 1, called the ground state)` corresponding to the number of loops in the standing wave use the model described above to answer the following there question for a particle moving in the line ` x = 0 to x = a Take h = 6.6 xx 10^(-34) J s and e = 1.6 xx 10^(-19)C` If the mass of the particle is `m = 1.0 xx 10^(-30) kg and a= 6.6 nm` the energyof the particle in its ground state is closest toA. `0.8 meV`B. `8 meV`C. `80 meV`D. `800 meV` |
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Answer» Correct Answer - B For ground state `n = 1`, Given `m = 1.0 xx 10^(-30)kg , a= 6.6 xx 10^(9) m ` `:. E = (t^(2) xx (6.6 xx 10^(-34)) ^(2))/(8 xx 1 xx 10^(-30) xx (6.6 xx 10^(-9))^(2)) 1 = 8 meV` |
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| 132. |
The `beta - decay` process , discoverwd around `1900` , is basically the decay of a neutron `(n)` in the laboratory , a proton `(p)` and an electron `(e^(bar))` are observed as the decay that the kinetic energy of the electron should be a constant . But experimentally , if was observed that the electron kinectic energy has continuous spectrum Considering a three- body decay process , i.e. ` n rarr p + e^(bar) + bar nu _(e) , ` around `1930` , pauli expained the observed `(bar nu_(e))` to be massaless and possessing negligible energy , and the neutrino to be at rest , momentum and energy conservation principle are applied from this calculate , the maximum kinectic energy of the electron is `0.8 xx 10^(6) eV` The kinectic energy carrect by the proton is only the recoil energy What is the maximum energy of the anti-neutrino ?A. zeroB. Mach less then `0.8 xx 10^(6)eV`C. Nearly `0.8 xx 10^(6)eV`D. Mach large then `0.8 xx 10^(6)eV` |
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Answer» Correct Answer - C The energy sharred between anti- neutrino and electron . If the energy of electron is almost zero then the maximum energy of anti - neutrino is rearly `0.8 xx 10^(6)eV` |
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| 133. |
The `beta - decay` process , discoverwd around `1900` , is basically the decay of a neutron `(n)` in the laboratory , a proton `(p)` and an electron `(e^(bar))` are observed as the decay that the kinetic energy of the electron should be a constant . But experimentally , if was observed that the electron kinectic energy has continuous spectrum Considering a three- body decay process , i.e. ` n rarr p + e^(bar) + bar nu _(e) , ` around `1930` , pauli expained the observed `(bar nu_(e))` to be massaless and possessing negligible energy , and the neutrino to be at rest , momentum and energy conservation principle are applied from this calculate , the maximum kinectic energy of the electron is `0.8 xx 10^(6) eV` The kinectic energy carrect by the proton is only the recoil energy If the - neutrono had a mass of `3 eV// c^(2)` (where c is the speed of light ) insend of zero mass , what should be the range of the kinectic energy `K.` of the electron ?A. `0 le k le 0.8 xx 10^(6) eV`B. `3.0eV le k le 0.8 xx 10^(6) eV`C. `3.0eV le k le 0.8 xx 10^(6) eV`D. `0 le k le 0.8 xx 10^(6) eV` |
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Answer» Correct Answer - D K should be less then `0.8 xx 10^(6) eV `as anti - neutrations will have same energy |
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| 134. |
`k_(a) ` wavelength emitted by an atom of atomic number `E= 11` is `lambda` find the atomic number for an atomic that amils `k_(a)` radiation with wavwlength `43`.A. `Z = 6`B. `Z = 4`C. `Z = 11`D. `Z = 44` |
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Answer» Correct Answer - A For K_(q) , (1)/(lambda) prop (Z- 1)^2) `From `(i), (lambda_(2))/(lambda_(1)) -((Z_(1) - 1)^(2))/((Z_(2) - 1) ^(2)) rArr ( 4lambda)/(lambda) = ((11 - 1)^(2))/((Z_(2) - 1)^(2))` `rArr Z_(2) - 1 = (10)/(2) rArr Z_(2) = 6` |
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| 135. |
In a photoelectric experiment anode potential is ploted againest plate currect. A. `A` and `B` will have different intensities while `B` and `C` will have different frequenciesB. `B` and `C` will have different intensities while `A` and `C` will have different frequenciesC. `A` and `B` will have different intensities while `A` and `C` will have different frequenciesD. `B` and `B` will have equal intensities while `A` and `B` will have different frequencies |
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Answer» Correct Answer - D From the graph it is clear that `A` and `B` have the same stopping potential and therefore, the sure frequency Also `B` and `C` have the same intensity |
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| 136. |
A proton has kinetic energy `E = 100 keV` which is equal to that of a photo is `lambda_(1)` . The ratio of `k_(2) // lambda_(1) ` is proportional toA. `E^(2)`B. `E^(1//2)`C. `E^(-1)`D. `E^(-1//2)` |
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Answer» Correct Answer - D For photon, `lambda_(2) = (hc)/(E )`…(i) For proton , P = sqrt(2mE)` `lambda_(1) = (h)/(P) = (h)/(sqrt(2mE)) ` …(ii) `(lambda_(2))/( lambda_(1)) = (hc)/(E xx (h)/(sqrt(2mE))) prop E^(1//2)` |
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| 137. |
After 280 days, the activity of a radioactive sample is 6000 dps. The activity reduces to 3000 dps after another 140 days. The initial activity of the sample in dps isA. (a) 6000B. (b) 9000C. (c) 3000D. (d) 24000 |
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Answer» Correct Answer - D Activity reduces from 6000 dps to 3000 dps in 140 days. It implies that half-life of the radioactive sample is 140 days. In 280 days (or two half-lives) activity will remain 1/4th of the initial activity. Hence, the initial activity of the sample is `4xx6000 dps=24000dps` Therefore, the correct option is (d). |
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| 138. |
The decay constants of a radioactive substance for a and b emission are `lambda_(a)` and `lambda_(b)` respectively. If the substance emits `alpha` and `beta` simultaneously, the average half life of the material will be -A. `lambda_(alpha) - lambda_(beta)`B. `lambda_(alpha) + lambda_(beta)`C. `(lambda_(alpha) lambda_(beta))/(lambda_(alpha) + lambda_(beta))`D. None of these |
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Answer» Correct Answer - C |
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| 139. |
What is the probability that a radioactive atom having a mean life of 10 days decays during the fifth day? |
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Answer» Correct Answer - C `1/lambda=10 days` `:.` `lambda=0.1day^-1` Probability of decay =Number of atoms decayed/Initial number of atoms `=(N_0(1-e^(-lambdat)))/(N_0)` `=1-e^(-lambdat)=1-e^(-0.1xx5)` `0.39` |
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| 140. |
The half-lives of radioisotypes `P^32` and `P^33` are 14 days and 25 days respectively. These radioisotopes are mixed in the ratio of 4:1 of their atoms. It the initial activity of the mixed sample is 3.0 mCi, find the activity of the mixed isotopes after 60 years. |
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Answer» Correct Answer - B::C `lambda_1=(1n2)/(T_1)` (T=half-life) `lambda_2=(1n2)/(T_2)` `R_(01)+R_0=8mCi` (given) `:.` `lambda_1(4N_0)+lambda_2(N_C)=8mCi` From here we can find number after `t=60 yr` `R=R_1+R_2` `=(4lambda_1N_0)e^(-lambda_1t)+(lambda_2N_0)e^(-lambda_2t)` |
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| 141. |
Atoms having the same …… but different ….. Are colled isotopes . |
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Answer» Correct Answer - A::B::C Atomic number , mass number |
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| 142. |
When `_(3)Li^(7)` nuclei are bombarded by protons , and the resultant nuclei are `_(4)Be^(8)` , the emitted particle will beA. alpha particleB. beta particleC. gamma particleD. neutrons |
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Answer» Correct Answer - C `_(3)^(7) Li + _(1)^(1)p rarr _(4)^(8) Be + _(0)^(0) gamma` |
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| 143. |
A radioactive ncleus A finaly transforms into a stable nucleus. B Then A and B can beA. isobarsB. isotonesC. isotopesD. None of these |
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Answer» Correct Answer - C |
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| 144. |
A radioactive material of half-life `T` was kept in a nuclear reactor at two different instants. The quantity kept second time was twice of the kept first time. If now their present activities are `A_1` and `A_2` respectively, then their age difference equalsA. `(T)/(ln2)|ln.(2A_(1))/(A_(2))|`B. `T|ln.(2A_(1))/(A_(2))|`C. `(T)/(ln2)|ln.(A_(1))/(2A_(2))|`D. `T|ln.(A_(2))/(2A_(1))|` |
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Answer» Correct Answer - C |
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| 145. |
Choose the correct options.A. (a) By gamma radiations atomic number is not changedB. (b) By gamma radiations mass number is not changedC. (c) By the emission of one `alpha` and two `beta` particles isotopes are producedD. (d) By the emission of one `alpha` and four `beta` particles isobars are produced |
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Answer» Correct Answer - A::B::C By the emission of an `alpha`-particle, atomic number decreases by 2 and by the emission of two particles atomic number increases by 2. Hence, net atomic number remains unchanged. |
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| 146. |
At `t=0`, number of radioactive nuclei of a radioactive substance are x and its radioactivity is y. Half-life of radioactive substance is T. Then,A. (a) `x/y` is constant throughoutB. (b) `x/ygtT`C. (c) value of `xy` remains half after one half-lifeD. (d) value of `xy` remains one fourth after one half-life |
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Answer» Correct Answer - A::B::D `y=lambdax=(1n2)/(T).x` `x/y=1/lambda=constant` `x/y=(T)/(1n2)` or `x/ygtT` (as `1n2=0.693)` Further, `xy=x(lambdax)=lambdax^2` After one half-life, x remains half. Hence, `x^2` remains `1/4th`. |
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| 147. |
Activity of a radioactive substance decreases from 8000 Bq to 1000 Bq in 9 days. What is the half-life and average life of the radioactive substance? |
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Answer» Correct Answer - A::B::C::D `1000=(1/2)^n8000` `:.` `n=3=number of half-lives` These half-lives are equivalent to 9 days. Hence, one half-life is 3 days. `t_(av)=1.44 t_(1//2)=1.44xx3=4.32days` |
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| 148. |
A radioactive substance contains `10^15` atoms and has an activity of `6.0xx10^11` Bq. What is its half-life? |
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Answer» Correct Answer - A::C `R_0=lambdaN_0` `6.0xx10^11=lambda(10^15)` `:.` `lambda=6.0xx10^(-4)s` `t_(1//2)=(1n2)/lambda=(0.693)/(6.0xx10^-4)` `=1.16xx10^3s` |
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| 149. |
What is the probability of a radioactive nucleus to survive one mean life?A. `(1)/("e")`B. `1-(1)/("e")`C. `(ln2)/("e")`D. `1-(ln2)/("e")` |
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Answer» Correct Answer - A |
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| 150. |
Atomic masses of two heavy atoms are `A_1` and `A_2`. Ratio of their respective nuclear densities will be approximatelyA. (a) `A_1/A_2`B. (b) `(A_1/A_2)^(1/3)`C. (c) `(A_2/A_1)^(1/3)`D. (d) `1` |
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Answer» Correct Answer - D Nuclear density is independent of A. It is of the order of `10^7 kg//m^3`. |
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