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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.
| 151. |
A radioactive element is disintegrating having half-life 6.93 s. The fractional change in number of nuclei of the radioactive element during 10 s isA. (a) 0.37B. (b) 0.63C. (c) 0.25D. (d) 0.50 |
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Answer» Correct Answer - B `lambda=(1n2)/(t_(1//2))=0.693/6.93=1/10sec^-1` `N=N_0e^(-lambdat)` `:.N/N_0=e^(-lambdat)=e^(-(1/10)(10))` `=e^-1~~0.63` |
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| 152. |
What is the probability of a radioactive nucleus to survive one mean life?A. (a) `1/e`B. (b) `(1)/(e+1)`C. (c) `1-1/e`D. (d) `1/e-1` |
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Answer» Correct Answer - A Probability of survival, P=Number of nuclei left/Initial number of nuclei`=(N_0e^(-lambdat))/(N_0)` At `t=`one mean life`=1/lambda` `P=e^-1=1/e` |
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| 153. |
A radioactive sample `S_1` having an activity of `5muCi` has twice the number of nuclei as another sample `S_2` which has an activity of `10muCi`. The half-lives of `S_1` and `S_2` can beA. (a) 20 yr and 5 yr, respectivelyB. (b) 20 yr and 10 yr, respectivelyC. (c) 10 yr eachD. (d) 5 yr each |
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Answer» Correct Answer - A Activity of `S_1=1/2` or `lambda_1N_1=1/2(lambda_2N_2)` or `lambda_1/lambda_2=(N_2)/(2N_1)` or `T_1/T_2=(2N_1)/(N_2)` (T=half-life`=(1n2)/(lambda)`) Given, `N_1=2N_2` `:.` `T_1/T_2=4` `:.` Correct option is (a). |
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| 154. |
`A` and `B` are isotopes. `B` and `C` are isobars. All three are radioactive. Which one of the following is true.A. A,b and C must belong to the same element.B. A,B and C may belong to the same element .C. It is possible that A will change to B though a radioactive-decay process.D. It is possible that B will change to C through a radioactive -decay process. |
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Answer» Correct Answer - d |
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| 155. |
Assertion : 1 amu is equal to 931.48 MeV. Reason: 1 amu is equal to 1/12th the mass of `C^12` atom.A. (a) If both Assertion and Reason are true and Reason is correct explanation of the Assertion.B. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. (c) If Assertion is true, but the Reason is false.D. (d) If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D (1 amu)(c^2)=931.48 MeV |
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| 156. |
The only source of energy in a particular star is the fusion reaction given by - `3._(2)He^(4)to._(6)C^(12)`+energy Masses of `._(2)He^(4)` and `._(6)C^(12)` are given `(m._(2)He^(4))=4.0025 u, m(._(6)C^(12))=12.0000u` Speed of light in vaccume is `3xx10^(8) m//s`. power output of star is `4.5xx10^(27)` watt. The rate at which the star burns helium isA. `8xx10^(12)`B. `4xx10^(12)`C. `12xx10^(13)`D. `9xx10^(13)` |
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Answer» Correct Answer - A |
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| 157. |
In the Bohr model of the hydrogen atgomA. the radius of the nth orbit is proportiona to `n_(2)`B. the total energy of the electron in the nth orbit is inversely proportional to nC. the angular momentum of the electron in an the orbit is an integral miltiple of `h//2pi`D. the magnitude of the potentital energy of the electron in any orbit is greater han its kinetic energy |
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Answer» Correct Answer - A::C::D |
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| 158. |
The mass number of a nucleus is.A. always less then it atomic numberB. always more than its atomic numberC. sometimes equal to its atomic numberD. sometimes more than its atomic number |
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Answer» Correct Answer - C::D |
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| 159. |
Light of wavelength `2000Å` is incident on a metal surface of work function 3.0 eV. Find the minimum and maximum kinetic energy of the photoelectrons. |
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Answer» Correct Answer - A::C `K_(min) = 0` and `K_(max) = E -W `=(12375)/(2000)- 3.0` `~~3.19 eV` |
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| 160. |
Fill in the blanks with appropriate items : Consider the following reaction, `.^(2)H_(1)+.^(2)H_(1)=.^(4)He_(2)+Q`. Mass of the deuterium atom= `2.0141 u ` , Mass of the helium atom = `4.0024 u` This is a nuclear ________ reaction in which the energy Q is released is ______ MeV. |
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Answer» Correct Answer - B::C::D This is a fusion reaction ` Energy released = (delta m) xx 931.5 MeV is ]` `= = [2 xx 2.0141 - 4.0024 ] xx 931.5 Mev` `= 24.03 MeV` |
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| 161. |
A radioactive sample `S_(1)` having an activity `5 mu Ci` has twice the number of inucle as another sample `S_(2)` which has as activity of `10 mu Ci` . The half lives of `S_(1) `and `S_(2) `can beA. `20 years and 5 years , respectively `B. `20 years and 10 years , respectively `C. `10 years eatch `D. `5 years eatch ` |
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Answer» Correct Answer - A Sample S- 1 Activity `5 nu Ci` No of nucle `N_(1) = 2N` `- ((dN)/(dt)0 _(1) = lambda_(1) N_(1)` rArr - 5 = lambda_(1) xx 2N ` ….(i) From (i) and (ii) `(5)/(10) = (lambda_(1) xx 2 N)/(lambda_(2) + N)` rArr (lambda_(1))/(lambda_(2)) = (1)/(4)` `rArr (T_(1//2)_(2))/((T_(1//2) _(1)) = (1)/(4) [(:. lambda prop (1)/(T_(1//2))]` Sample S - 2 `10 mu Ci` `N_(2) = N` `- ((dN)/(dt)) _(2) = lambda_(2) N_(2)` ` - 10 = lambda_(2) prop N` ....(ii) ` |
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| 162. |
A potential diffrence , the `20kv` is applied across an X- ray s tube . The minimum wavelength of x- ray generated is …….` Å` . |
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Answer» Correct Answer - B `lambda_(min) = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(1.6 xx 10^(-19) xx 20 xx 10^(2)) = 0.62 Å` |
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| 163. |
In a hypothetical atom, mass of electron is doubled, value of atomic number is `Z = 4`. Find wavelength of photon when this electron jumps from 3rd excited state to 2nd orbit. |
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Answer» Correct Answer - A `E prop (Z^2)/(n^2)`prop m` Mass is doubled, `Z = 4` and 3rd excited state means `n =4`, second orbit means `n = 2` for these` values, we have. `E_4 =- 13.6xx2((4)/(4))^2 =- 27.2eV` and `E_2 =-13.6xx2((4)/(2))^2 =- 108.8eV` `lmabda(inÅ)= (12375)/(DeltaE(in eV)) = (12375)/(E_4 - E_2)` `=(12375)/(-27.2+108.8)` `:. lambda = 151.65Å` |
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| 164. |
A nucleus with `Z = 92` emits the following in a sequence `a, beta^(bar) , beta^(bar)a,a,a,a,a, beta^(bar) , beta^(bar) , a, beta^(+) , beta^(+) , a ` Them `Z` of the resulting nucleus isA. `76`B. `78`C. `82`D. `74` |
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Answer» Correct Answer - B The number of `a - particle released = 8` Therefore the atomic number should decreases by `16` The number of `beta^(bar) - particle released = 4 ` Therefore the atomic number should increases by `4` Also the number of `beta ^(+)` particle released is `2` which should decreases the atomic number is `92 - 16 + 4 - 2 = 78` |
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| 165. |
In problem` 43`, number of atoms decayed between time interval `t_(1)` and `t_(2)` areA. `(ln(2))/(lambda)(R_(1)-R_(2))`B. `R_(1)e^(-lambdat_(1))-R_(2)e^(-lambdat_(2))`C. `lambda(R_(1)-R_(2))`D. `((R_(1)-R_(2))/(lambda))` |
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Answer» Correct Answer - D |
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| 166. |
The energy of a hydrogen atom in its ground state is `-13.6 eV`. The energy of the level corresponding to the quantum number n=5 isA. `-0.54 eV`B. `-5.40 eV`C. `-0.85 eV`D. `-2.72 eV` |
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Answer» Correct Answer - A `E prop (1)/(n^2)` `:. E_5 =(-13.6)/((5))^2` =-0.544 eV. |
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| 167. |
The magnitude of angular momentum, orbital radius and time period of revolution of an electron in a hydrogen atom corresponding to the quantum number n are L , r and T respectively. Which of the following statement (s) is/are correct?A. `(rL)/T` is independent of nB. `L/T prop 1/(n^2)`C. `T/r prop n`D. `Lr prop 1/n^3` |
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Answer» Correct Answer - A::B::C `L prop n, r prop n^2` and T =(2pir)/(v) or T prop (r )/(v) prop (n^2)/((1//n))` or `T prop n^3` |
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| 168. |
In which of the following cases the havier of the two particles has a smaller de-Broglie wavelength ? The two particlesA. move with the same speedB. move with the same linear momentumC. move with the same kinetic energyD. have the same change of potential energy in a conservative field |
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Answer» Correct Answer - A::C `lambda = (h)/(p) = (h)/(mv) = (h)/(sqrt2Km)` `lambda prop(1)/(m) (if v is same) `lambda prop (1)/(sqrtm) (if K is same) If change in potential energy is same, then change in kinetic energy is also same. But , this does not mean the kinetic energy is same. |
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| 169. |
Find the energy and mometum of a photon of ultraviolet radiation of 280 nm wavelength. |
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Answer» Correct Answer - A::B::D `E = (hc)/(lambda) , P =(h)/(lambda) |
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| 170. |
For uranium nucleus low does its mass very with volume?A. `m prop V`B. `m prop 1//V`C. `m prop sqrt(V)`D. `m prop 1^(/2) |
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Answer» Correct Answer - A KEY CONCEPT : we know that radius of the nucleus `R = R_(0)A^(1//3)` , where `A` is the mass number `:. R^(2) = R_(0)^(3) A` `rArr (4)/(3) pi R^(3) = (4)/(3) pi R_(0)^(3) A rArr volume prop mass` |
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| 171. |
A nucles with mass number `220`initilly at rest enits an `a - particle`. If the `Q` value of the reaction is `5.5MeV`, calculate the kinetic energy of the `a - particle`A. `4.4 MeV`B. `5.4 MeV`C. `5.6 MeV`D. `6.5 MeV` |
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Answer» Correct Answer - B By conservation of momentum , `p_(1) = p_(2) ` ` sqrt(2K_(1) m_(1)) = sqrt(2K_(2) m_(2)) ` `rArr sqrt(2K_(1)(216)) = sqrt(2K_(2)(4)) ` rArr K_(2) = 54K_(1) ` …(i) `Also , K_(1) + K_(2) = 5.5 MeV` …(ii) solve equation (i) and (ii) |
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| 172. |
A proton is first from very loward a nucleus with charge `Q = 120 e , ` where e is the nucles The de Brogle wavelength (in unit of fin) of the proton at its start is (tke the proton mass , `m _(p) = (5//3) xx 10^(-27) kg h//s = 4.2 xx 10^(-15) J s // C,` `(1)/(4 pi s_(0)) = 9 xx 10^(9) m//F , 1 fm = 10^(-15) m ` |
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Answer» `1` cos in K.E. of proton = gain in potential energy of the proton - nucleus system `(1)/(2) m nu^(2) = (1)/(4 pi s_(0)) (q_(1) q_(2))/(r )` `:. (p^(2))/(2m) = (1)/(4 pi epsilon_(0))(q_(1) q_(2))/(r )` :. (1)/(2m)((h^(2))/(h^(2))) = (1)/(4 pi epsilon_(0))(q_(1) q_(2))/(r )` `:. lambda = sqrt((4 pi epsilon_(0) r , h^(2))/(q_(1) q_(2) (2m))) = 7 fm` |
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| 173. |
A silver of radius `1 cm` and work function `4.7 eV` is suspended from an insulating thread in freepace it is under continuous illumination of `200 nm` wavelength light AS photoelectron are emitted the sphere gas charged and acquired a potential . The maximum number of photoelectron emitted from the sphere is `A xx 10^(e) (where 1lt A lt 10)` The value of `z` is |
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Answer» Stopping potential `= (1)/(e ) [(hc)/(lambda) - phi]` where `hc = 1240 eV - nm` `= (1)/(e ) [(1240)/(200) - 4.7] = ((1)/€ [6.2 - 4.7]` `= (1)/€ xx 1.5 eV = 1.5 eV` `But V = (1)/(4 pi s_(0)) (q)/® = (1)/(4 pi s_(0)) (ne)/®` `:. N = (Vr(4 pi s_(0)))/(e) = (1.5 xx 10^(-2))/(9 xx 10^(9) xx 1.6 xx 10^(-19))` Compering it with `A xx 10^(2)` we get `z= 7` |
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| 174. |
When fission occurs, several neutrons are released are released and the fission fragements are beta radioactive, why? |
| Answer» Correct Answer - A::B::C::D | |
| 175. |
The work function of Na metal is `2.3 eV`, then the threshold wavelength lies in the following region of EM spectrum -A. UltravioletB. X-rayC. VioletD. Yellow |
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Answer» Correct Answer - A |
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| 176. |
The patient is asked to drink `BaSO_(4)` for examining the stomach by X-rays because X-rays are-A. Reflected by heavy atomsB. Refracted by heavy atomsC. Less absorbed by heavy atomsD. More absorbed by heavy atoms |
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Answer» Correct Answer - D |
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| 177. |
When a `beta^(-)` particle is emitted from a nucleus, the neutrons-proton ratio:A. os decreasedB. is increasedC. remains the sameD. may in increases or decreases |
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Answer» Correct Answer - A |
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| 178. |
An electron has a de Broglie wavelength of `2.80xx10^(-10)` m. Determine (a) the magnitude of its momentum, (b) its kinetci energy(in joule and in electron volt). |
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Answer» Correct Answer - A::B::C::D (a) `lambda rArr (h)/(P) rArr p=(h)/(lambda)` (b) `lambda = (lambda)/(sqrt2km) rArr K = (h^2)/(2m lambda^2).` |
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| 179. |
(a) An electron moves with a speed of `4.70xx10^6` m//s. What is its de-Broglie wavelength? (b) A proton moves with the same speed. Determine its de - Broglie wavelength. |
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Answer» Correct Answer - A::B::D `lambda = (h)/(mv)` |
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| 180. |
If `I_(0)` and `I_(a)` denote intensities of incident and absorbed X-rays, then-A. `I_(a) = I_(0) e^(-mu d)`B. `I_(a) = I_(0) (1 - e^(-mu d))`C. `I_(a) = I_(0) (1 - e^(mu d))`D. `I_(a) = I_(0) e^(mu d)` |
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Answer» Correct Answer - A |
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| 181. |
Determine current I in the configuration- A. 1 ampB. 0 ampC. less than 1 ampD. None |
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Answer» Correct Answer - A |
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| 182. |
Generation of X-rays is a-A. Phenomenon of conversion of KE into radiant energyB. Principle of conservation of momentumC. Phenomenon of conversion of mass into energyD. Principle of conservation of electric charge |
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Answer» Correct Answer - A |
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| 183. |
Following diagram shows relation between emitted X-rays intensity and wavelength obtained X-ray tube its sharp peaks A and B shows A. band spectrumB. continuous spectrumC. characteristics radiationsD. white radiations |
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Answer» Correct Answer - A |
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| 184. |
Radiation two photons having energies twice and five times the work function of a metal are incident successively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases will beA. `1:2`B. `2:1`C. `1:4`D. `4:1` |
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Answer» Correct Answer - A |
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| 185. |
Two identical capacitors A and B are charged to the same potential V and are connected in two circuits at `t = 0` as shown in figure. The charge of the capacitors at a time `t = CR` are respectively- A. VC, VCB. `VC//e, VC`C. `VC, VC//e`D. `VC//e, VC//e` |
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Answer» Correct Answer - A |
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| 186. |
The particles emitted in the nuclear reaction are respectively - `._(Z)X^(A) rarr ._(Z + A)Y^(A) rarr._(Z - 1)R^(A - 4) rarr._(Z - 1)R^(A - 4)`A. `beta, gamma, alpha`B. `alpha, beta, gamma`C. `beta, alpha, gamma`D. `gamma, alpha, beta` |
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Answer» Correct Answer - A |
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| 187. |
The order of magnitude of the density of nuclear matter is `10^(4) kg m^(-3)` |
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Answer» Density `= (m)/(V) = (A xx 1.67 xx 10^(-27))/((3)/(4) pi [R_(0) A^(1//3)]^(3)` `= (1.67 xx 10^(-27))/(1.33 xx 3.14 xx (1.1 xx 10^(-15))) = 3 xx 10^(17) kg//m^(3)` |
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| 188. |
A radioative sample emit `n beta` - particle is `2` sec , in next `5 eV` ` sec it emit ` 0.75 n beta`- particle , what is the mean life of the sample? |
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Answer» Correct Answer - B::C::D `lambda = (log_(e)(A_(0))/(A)/(t) = (1)/(2) log_(e) (n)/(0.75n)` `rArr Mean Life = (1)/(lambda) = (2)/(log_(e) 4 //3)` |
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| 189. |
Excitions energy of hydrogen atom is `13.6 eV` mathc the following |
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Answer» Correct Answer - A::B::C |
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| 190. |
The plate resistance of a triode in `3 xx 10^(3)` aloms and its muthal of the triode isA. ` 5 xx 10^(-5) `B. ` 4.5`C. `45`D. `(0.2 xx10^(3)` |
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Answer» Correct Answer - B (b) KEY CONCEPTWe know that `mu= g_(m) xx r_(0)` where `nu= `amplification factor, `g_(m) = `mutual conductance `r_(0) = `plate resistance `:. Mu = 3 xx 10^(3) xx 1.5 xx 10^(-3) = 4.5` |
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| 191. |
In a ……. Biassed `p- alpha` of janction , the net flow of holes is from the `n` region to the `p` regain . |
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Answer» Correct Answer - A Reverse |
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| 192. |
In the nuclear process , `_(6) C^(11) rarr _(2)B^(11) + beta^(11) beta^(+) + X , X` stadnds for……. |
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Answer» `_(6)^(11) C rarr _(3)^(11) B + beta ^(+) + X rArr _(8)^(11)C rarr _(5)^(11) B + _(+) ^(0) e + v ` (nutrino) The batancing of atomic number and mass number is correct . Therefore , `X` stands for neutrion . |
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| 193. |
In a nuclear reactor mathc the following . |
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Answer» Correct Answer - A::B::C::D |
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| 194. |
The half life of radioactive Radon is `3.8 days` . The time at the end of which `(1)/(20) th` of the radon sample will remain undecayed is `(given log e = 0.4343 ) `A. `3.8 days`B. `16.5 days`C. `33 days`D. ` 76 days` |
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Answer» Correct Answer - B (b) t_(1//2) = 3.8 `days `:. Lambda = (0.693)/(t_(1//2)) = (0.693)/(3.8) = 0.182` If the initial number of aton is `a = A_(0)` then after time `t` the number of aloms is `20 = A` . We have to find `t` `t = (2.303)/(lambda) log (A_(0))/(A) = (2.303)/(0.182) log (a)/(a//20) = 16.46 days` |
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| 195. |
If element with particle quantum number `ngt4` were not allowed in nature , the number of posible elemant would beA. `60 `B. `32`C. `4`D. `64` |
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Answer» Correct Answer - A (a) KEY CONCEPT : The maximum number of electrons in an orbit is `2 n^(2) , ngt 4 ` is not alllowed. therefore the number of maximum electron that can be in first four orbit are `2 (1)^(2) + 2 (3)^(2) + 2 (4)^(2) ` `= 2 + 8 + 18 + 32 = 60 ` Therefore , possible element are `60` . |
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| 196. |
When npn transistor is used as an ampliflerA. electron move from collector to baseB. boles move from emitted to baseC. electron move from base to collectorD. holes move from base to emiter |
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Answer» Correct Answer - C Electrons move from base to emmitter |
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| 197. |
The graph of `1n (R/R_0)` versus `1n A (R = radius` of a nucleus and `A =` its mass number) isA. a straight lineB. a parabolaC. and ellipseD. a circle |
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Answer» Correct Answer - A |
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| 198. |
The mass number of a nucleus isA. always less then its number isB. always more then its number isC. sometimes equal to its atomic numberD. sometimes more then and sometimes equal to its atomic number |
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Answer» Correct Answer - C::D In the case of is hydrogen , atomic = mass number in the atoms atomic number lt mass number |
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| 199. |
When an electron moving at a high speed strikes a metal surface, which of the following are possible? (i) The entire energy of the electron may be converted into an X-ray photon (ii) Any fraction of energy of the electron may be converted into an X-ray photon (iii) The entire energy of the electron may get converted to heat (iv) The electron may undergo elastic collision with the metal surfaceA. The entire energy of the electron may be converted into an X-ray photon .B. Any fraction of the energy of the electron may be converted into an X-ray phopton.C. The entire energy of the electron may get converted to heat.D. The electron may undergo elastic collision with the metal surface. |
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Answer» Correct Answer - a,b,c,d |
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| 200. |
The X- rays beam coming from an X- rays tube will beA. monuchromaticB. having all wavwlength smaller than a cortain maximum wavelengthC. having all wavwlength largest than a cortain maximum wavelengthD. having all wavwlength lying between a minimum and a maximum wavelength |
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Answer» Correct Answer - C in the correct option `lambda_(min)= (hc)/(eV)` |
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