419 + Interview Questions in MODERN-PHYSICS IN GENERAL AWARENESS Page 2 InterviewSolution

51.	A sample of radioactive substance loses half of its activity in 4 days. The time in which its activity is reduced to 5% isA. (a) 12 daysB. (b) 8.3 daysC. (c) 17.3 daysD. (d) None of these
Answer» Correct Answer - C `lambda=(1n2)/(t_(1//2)) = (1n2)/(4) day^-1` Now, apply `R=R_0e^(-lambdat) implies 5=100e^(-lambdat)` Substituting value of `lambda`, we can find t.

Discussion

52.	The activity of a radioactive sample goes down to about 6% in a time of 2 hour. The half-life of the sample in minute is aboutA. (a) 30B. (b) 15C. (c) 60D. (d) 120
Answer» Correct Answer - A Activity of atoms is 6.25% after four half-lives. `:.` Four half-lives`~~2h=120min` `:.` One half-life is 30 min.

Discussion

53.	A radioactive substance has a half-life of 64.8 h. A sample containing this isotope has an initial activity `(t=0)` of `40muCi`. Calculate the number of nuclei that decay in the time interval between `t_1=10.0h` and `t_2=12.0h`.
Answer» Correct Answer - A::C::D `R_(0)=lambdaN_(0)impliesN_(0)=(R_(0))/(lambda)` where, `lambda=(In2)/(t_(1//2))` `N=N_(0^(e-lambdat))` Find `N_(1)=N_(0e^(-lambdat1))` and `N_(2)=N_(0e^(-lambdat2))` `:.` Number of nuclei decayed in given time `=N_(1)-N_(2)`

Discussion

54.	A radioactive sample contains `1.00xx10^15` atoms and has an activity of `6.00xx10^11` Bq. What is its half-life?
Answer» Correct Answer - A::B `R=lambdaN` `6xx10^11=1.0xx10^15 lambda` `:.` `lambda=6xx10^-4s` `t_(1//2)=(1n2)/(lambda)=(0.693)/(6xx10^-4)s` `=1155s=19.25` min

Discussion

55.	At time `t=0`, activity of a radioactive substance is 1600 Bq, at t=8 s activity remains 100 Bq. Find the activity at t=2 s.
Answer» Correct Answer - D `R=R_0(1/2)^n` Here , n is the number of half-lives. Given, `R=(R_0)/(16)` :. `R_0/16 = R_0(1/2)^n` or `n=4` Four half-lives are equivalent to 8 s. Hence, 2 s is equal to one half-life. So, in one half-life activity will remain half of `1600 Bq`, i.e. `800 Bq`.

Discussion

56.	At time `t=0` , number of nuclei of a radioactive substance are 100. At `t=1` s these numbers become 90. Find the number of nuclei at `t=2 s`.
Answer» Correct Answer - A In 1 second, 90% of the nuclei have remained undecayed, so in another 1 second 90% of 90, i.e. 81 nuclei will remain undecayed.

Discussion

57.	Two electrons are moving with the same speed, v One electron enters at different instants , the quantity produced second time was twic of that produced first time. If now their present acitivies are `A_(1)` and `A_(2)` respectively then their age difference equalsA. `lambda_(1)=lambda_(2)`B. `lambda_(1)>lambda_(2)`C. `lambda_(1) < lambda_(2)`D. `lambda_(1) > lambda_(2)` or `lambda_(1) < lambda_(2)`
Answer» Correct Answer - D

Discussion

58.	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. (a) `(T)/(1n2)1n(2A_1)/(A_2)`B. (b) `T1n(A_1)/(A_2)`C. (c) `(T)/(1n2)1n(A_2)/(2A_1)`D. (d) `T1n(A_2)/(2A_1)`
Answer» Correct Answer - C Activity `Aprop` Number of atoms` `A_1=A_0e^(-lambdat_1)` `:.` `t_1=1/lambda1n((A_0)/(A_1))=(T)/(1n2)1n(A_0)/(A_1)` `A_2=2A_0e^(-lambda_(t_2))` `t_2=(T)/(1n2)1n((2A_0)/(A_2))` `t_1-t_2=(T)/(1n2)(A_0/A_1xx(A_2)/(2A_0))` `=(T)/(1n2)1n((A_2)/(2A_1))`

Discussion

59.	In a certain nuclear reactor, a radioactive nucleus is bieng produced at a constant rate =1000/s The mean life of the radionuclide is 40 minutes. At steady state, the number of radionuclide will beA. `4xx10^(4)`B. `24xx10^(4)`C. `24xx10^(5)`D. `24xx10^(6)`
Answer» Correct Answer - C

Discussion

60.	The combination of shown below yiedsA. `OR gate`B. `NOT gate`C. `XOR gate`D. `NAND gate`
Answer» Correct Answer - A The final doolean expression is `X =bar((bar(A). bar(B))) = bar (barA) + bar (barB ) = A + B rArr OR gate

Discussion

61.	A nucleus `._(Z)X^(A)` emits 2 `alpha -` particles and 3 `beta -` particles. The ratio of total protons and neutrons in the final nucleus is :A. `(Z - 7)/(A - Z + 7)`B. `(Z - 1)/(A - Z - 8)`C. `(Z - 1)/(A - Z - 7)`D. `(Z - 3)/(A - Z + 3)`
Answer» Correct Answer - C

Discussion

62.	This question contains Statement - 1 and Statement -2 Of the four choice given after the Statements , choose the one that best decribes the two Statements Statement- 1: Energy is reased when heavy underge fission or light nuclei undergo fasion and Statement- 2: for nuclei , binding energy nucleon increases with increasing `Z` while for light nuclei it decreases with increasing `Z`A. Statement - 1 is false ,Statement - 2 is trueB. Statement - 1 is true ,Statement - 2 is true, Statement - 2 is a correct explanation for Statement - 1C. Statement - 1 is true ,Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1D. Statement - 1 is true ,Statement - 2 is false
Answer» Correct Answer - D we know that energy is released when heavy nuclei undergo fission or light nuclei undergo fission Therefore statement (1) is correct . The second statement is false because for heavy nuclei the bimding energy per nucleon decrease with increase `Z `and for light nuclei B.E nucleon increase with increasing `Z`

Discussion

63.	Stopping potential of 24, 100, 110 and 115 k V are measured for photoelectrons emitted from a certain element when it is radiated with monochromatic X-ray . If this element is used as a target in an X-ray tube,what will be the wavelength of `K_a` - line?
Answer» Correct Answer - A::C Stopping potentials are 24, 100, 110 and 115 kV, i.e. if the electrons are emitted from conduction band maximum kinetic energy of photoelectrons would be `115xx10^3` eV. If they are emitted from next inner shell, maximum kinetic energy of photoelectrons would be `110xx10^3 eV` and so on. for photoelectrons of L- shell it would be `100xx10^3` eV and for K- shell it is`24xx10^3` ev. Therefore, diference between energy of L-shell and K-shell is `Delta E = E_L - K_K` `=(100 - 24)xx10^3 eV` `= 76xx10^3`eV `:.` Wavelength of `K_a` line (transition of electron form L-shell to K-shell) is, `lambda_(ka) (in Å) = (12375)/(DeltaE(in eV) = (12375)/(76xx10^3)` `= 0.163 Å`.

Discussion

64.	What will be the maximum kinetic energy of the photoelectrons ejected from magnesium (for which the work -function W=3.7 eV) when irradiated by ultraviolet light of frequency `1.5xx10^15 s^(-1)`.
Answer» Correct Answer - A::B `E = hf=((6.63xx10^(-34)(1.5 xx10^(15)))/(1.6 xx10^(-19)` `K_(max) = E - W = 6.21 - 3.7 = 2.51 eV.

Discussion

65.	The stopping potential for the photoelectrons emitted from a metal surface of work function 1.7 eV is 10.4 V. Find the wavelength of the radiaiton used. Also, identify the energy levels in hydrogen atom, which will emit this wavelength.
Answer» Correct Answer - A::B::C Stopping potential =10.4 V `:. K_(max) = 10.4 eV` `E = W + K_(max) = 1.7+ 10.4 = 12.1 eV` `lambda = (12375)/(12.1) = 1022 Å.` 12.1 eV is the energy gap between n = 3 and n=1 in hydrogen atom.

Discussion

66.	Light of wavelength `0.6mum` from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0.5V. With light of wavelength `0.4mum` from a murcury vapor lamp, the stopping potential is `1.5V`. Then, the work function [in electron volts] of the photocell surface is
Answer» Correct Answer - 4

Discussion

67.	In a photeclectric experiment , when electromegnetic wave given by `E=E_(0)"sin"omegat` is incident , electron just ejects. When `E=E_(0)"sin"2omegat` is incident `K_("max")=K_(1)` and when E`=E_(0)"sin" 6omegat` is incident `K_("amx")=k_(2)`. When `k_(2)//k_(1)`
Answer» Correct Answer - 5

Discussion

68.	The radiation corresponding to ` 3 rarr 2` transition of hydrogen atom falls on a metal surface to produce photoelectrons . These electrons are made to enter circuit a magnitic field `3 xx 10^(-4) T` if the ratio of thelargest circular path follow by these electron is `10.0 mm , the work function of the metal is close toA. `1.8 eV`B. `1.1 eV`C. `0.8 eV`D. `1.6 eV`
Answer» Correct Answer - B Radius of circular path followedb by electron is given by `r = (m v)/(qB) = (sqrt(2m e V))/(eB) = (1)/® (sqrt(2m))/€V` rArr V = (n^(2) r^(2) e)/(2m) = 0.W` for transition between `3 to 2` `E = 13.6 ((1)/(4) - (1)/(9)) = (13.6 xx 5)/(36) = 1.88 eV` work function `= 1.88 eV - 0.8 eV` `= 1.08 eV = 1.1 eV`

Discussion

69.	If a star can convert all the He nuclei completely into oxygen nuclei. The energy released per oxygen nuclei is (Mass of the helium nucleus is 4.0026 amu and mass of oxygen nucleus is 15.9994 amu)A. (a) 7.6 MeVB. (b) 56.12 MeVC. (c) 10.24 MeVD. (d) 23.4 MeV
Answer» Correct Answer - C `4(`_2He^4)=`_8O^16` Mass defect, `Deltam={4(4.0026)-15.9994}` `=0.011` amu `:.` Energy released per oxygen nuclei `=(0.011)(931.48)MeV` `=10.24` MeV `:.` Correct answer is (c).

Discussion

70.	At a specific instant emission of radioactive compound is deffernt in a field . The compound can emit (i) electron (ii)protons(iii)`He^(2+ )`(iv) neutrons The emission at instant can beA. I,ii,iiiB. I,ii,iii,ivC. ivD. ii,iii
Answer» Correct Answer - A charged particles are deflected in magnatic field

Discussion

71.	A small quantity of solution containing `Ne^(24)` radio nucliode (half life `= 15 hour`) of activity `1.0` microcurlar is injected into the blood of a person A sample of the blood of volume `1 cm^(3)` taken a after `5` hour shown an activity of the blood in the body of the person . Assume that redicative solution mixed uniformly in the blood of the person `(1 curie = 3.7 xx 10^(10)` disntegrations per sound)
Answer» `t_(1//2) = 15 hour` Actually `A_(0) = 10^(-6)` curie (in small quentyty of solution of `^(24) Na) = 3.7 xx 10^(4) dps` Observation of blood of volume `1 cm^(3)` After `5 hour .,A = 296 dps` The initial activity can be found by the formula ` t = (2.303)/(lambda) log_(10) (A_(0))/(A) rArr 5 = (2.303)/(0.693//15) xx log_(10) (A_(0))/(296)` `rArr log_(10) (A_(0))/(296)= ( 5 xx 0.693)/(2.303 xx 15) = (0.33010)/(3) = 0.10033` `rArr (A_(0))/(296)= 1.26 rArr A_(0) = 373 dpm = (373)/(60) dps` This is the activity level in `1 cm^(3)` Comparing it with the initial activity level of `3.7 xx 10^(4) dps `we find the volume of blood `V = (3.7 xx 10^(4))/(373//60) = 5951.7 cm^(3) = 5.951 litre `

Discussion

72.	A hydrogen like atom (atomic number Z) is uin a higher excleted atate of quantum n , The excited atom can make a two photon of energy `10.2 and 17.0 eV` respactively , Alernately the atom from the same excited state by successively eniting two photons of energies `4.25 eV and 5.95 eV` respectively Determine the value of n and Z (lonization energy of H- atom `= 13.6 eV`)
Answer» Correct Answer - C For hydrogen like atoms E_(0) = (13.6)/(n^(2)) Z^(2) eV//atom` `Given E_(n) - E_(2) = 10.2 + 17 = 27.2 eV` ….(i) `E_(n) - E_(3) = 4.24 + 5.95 = 10.2 eV` `:. E_(3) - E_(2) = 17` `But E_(3) - E_(2) = (13.6)/(9)Z^(2) - ((-13.6)/(4)Z^(2))` `= - 13.6 Z^(2) [(1)/(9) - (1)/(4)]` `= - 13.6 Z^(2) [(4 - 9)/(36)] = (13.6 xx 5)/(36)Z^(2)` ` (13.6 xx 5)/(36)Z^(2) = 17 rArr Z= 3 `E_(n) - E_(2) = - (13.6)/(n^(2)) xx 3^(2) - [-(13.6)/(2^(2) xx 3^(2) ]` `- 13.6 [(9)/(n^(2)) - (9)/(4)] = -13.6 xx 9 [(4 - n^(2))/(4 n^(2))]` ....(ii) frequency (i) and (ii) ` -13.6 xx 9 [(4 - n^(2))/(4 n^(2))]= 27.2` `rArr 122.4 (4 - n^(2)) = 108.8 n^(2)` `rArr n^(2)= (489)/(13.6) = 36 rArr n = 6`

Discussion

73.	An electron is excited from a lower energy state to a higher energy state in a hydrogen atom. Which of the following quantity/quantities decreases/decrease in the excitation ?A. potential energyB. Angular speedC. Kinetic energyD. Angularn momentum
Answer» Correct Answer - B::C

Discussion

74.	An electron of kinetic enregy K collides elastically with a stationary hydrogen atom is the ground state. Then,A. `K gt 13.6 eV`B. `K gt 10.2 eV`C. `K lt 10.2 eV`D. data insufficient
Answer» Correct Answer - C For `Kge10.2 eV` electrons can excite the hydrogen atom `(as E_2 - E_10.2 eV).` So, collision may be inelastic.

Discussion

75.	Assertion : `gamma`-rays are produced by the transition of a nucleus from some higher energy state to some lower energy state. Reason : Electromagnetic waves are always produced by the transition process.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.
Answer» Correct Answer - C In moving from lower energy state to higher energy state electromagnetic waves are absorbed.

Discussion

76.	In a hydrogen atom, the binding energy of the electron in the ground state is `E_(1)` then the frequency of revolution of the electron in the nth orbit isA. `2E_(1)//nh`B. `2E_(1)n//h`C. `E_(1)//nh`D. `E_(1)n//h`
Answer» Correct Answer - A

Discussion

77.	Figure represent some of the lower energy level of the hydrogen atom in simplified from. If the transition of an electron from `E_(4) to E_(2)` were accociated with the emission of blue light , which one of the following transition could be accociated with the emission of red light? A. `E_(4) " to " E_(1)`B. `E_(3)" to " E_(2)`C. `E_(2) " to " E_(3)`D. `E_(1)" to "E_(4)`
Answer» Correct Answer - A

Discussion

78.	Nuclei X and Y convert into a stable nucleus Z. At t=0 , the number of nuclei of X is 8 times that of Y. Half life of X is 1 hour and half life of y=2 hour. Find the time (in hour) at which rate of disintegration of X and Y are equal.
Answer» Correct Answer - 8

Discussion

79.	Regarding a nucleus choose the correct options.A. (a) Density of a nucleus is directly proportional to mass number AB. (b) Density of all the nuclei is almost constant of the order of `10^17kg//m^3`C. (c) Nucleus radius is of the order of `10^-15m`D. (d) Nucleus radius `prop` A
Answer» Correct Answer - B::C `R=R_0A^(1//3)` or `RpropA^(1//3)`

Discussion

80.	The atomic masses of the hydrogen isotopes are Hydrogen `m_1H^1=1.007825` amu Deuterium `m_1H^2=2.014102` amu Tritium `m_1H^3=3.016049` amu The mass of deuterium, `_1H^2` that would be needed to generate 1 kWhA. (a) `3.7kg`B. (b) `3.7g`C. (c) `3.7xx10^-5kg`D. (d) `3.7xx10^-8kg`
Answer» Correct Answer - D In one fusion reaction two `_1^2H` nuclei are used. Hence, total number of `_1^2H` nuclei are 2N. or `1.125xx10^19` Mass in kg `=((1.125xx10^19)/(6.02xx10^26))(2)kg` `=3.7xx10^-8kg`

Discussion

81.	The atomic masses of the hydrogen isotopes are Hydrogen `m_1H^1=1.007825` amu Deuterium `m_1H^2=2.014102` amu Tritium `m_1H^3=3.016049` amu The number of fusion reactions required to generate 1kWh is nearlyA. (a) `10^8`B. (b) `10^18`C. (c) `10^28`D. (d) `10^38`
Answer» Correct Answer - B Let N number of fusion reactions are required, then `Nxx4xx1.6xx10^-13=10^3xx3600` `N=5.625xx10^18`

Discussion

82.	The atomic masses of the hydrogen isotopes are Hydrogen `m_1H^1=1.007825` amu Deuterium `m_1H^2=2.014102` amu Tritium `m_1H^3=3.016049` amu The energy released in the reaction, `_1H^2+_1H^2rarr_1H^3+_1H^1` is nearlyA. (a) 1MeVB. (b) 2MeVC. (c) 4MeVD. (d) 8 MeV
Answer» Correct Answer - C Energy released `=(Deltam)(931.48)MeV` `=[2xx2.01102-3.0160-1.007825]xx931.5` `=4.03MeV=4MeV`

Discussion

83.	For a certain hypothetical one electron atom, the wavelength `(in Å)` for the spectral lines for transitions originating at n=p and terminating at n=1 are given by `lambda = (1500 p^2)/(p^2 - 1), where p = 2,3,4` (a)Find the wavelength of the least energetic and the most energetic photons in this series.(b) Construct an energy level diagram for this element showing the energies of the lowest three levels. (c ) What is the ionization potential fo this elelment?
Answer» Correct Answer - A::B::C::D (a) `lambda =1500 ((1)/(1 -1/p^2))` `lambda_(max)` corresponds to least energetic photon with p = 2. `:. lambda_(max) = 1500((1)/(1-1//4)) = 2000 Å` `lambda_(min)` corresponds ot most energetic photon `with p = 00` `lambda_(min) = 1500 Å` (b) `lambda_(oo -1) =1500 Å` `...........E_3 = - 0.95 eV` `........... E_2 = -2.05 eV` `........... E_1 = -8.25 eV` `:. E_(oo) - E_1 = (12375)/(1500) eV` =8.25 eV `:. E_1 = - 8.25 eV (as E_(oo) = 0)` `lambda_(2 -1) = 2000 Å` `:. E_2 -E_1 = (12375)/(2000) eV`h =6.2 eV `:. E_2 = -2.05 eV` Similarly, `lambda_(31) = 1500((1)/(1-1//9))` =1687.5 Å `:. E_3 - E_1 = (12375)/(1687.5) eV = 7.3 eV` `:. E_3 = - 0.95 eV` (c ) lonization potential = 8.25 V.

Discussion

84.	when a beam of `10.6 eV` photons of intensity `2.0 W//m^(2)` falls on a platinum surface of area `1.0 xx 10^(4) m^(2)` and work function `5.6 eV , 0.53 %` of the incidentphotons eject photoelectrons find the number of photoelectrons emited per second and their minimum energies (in eV)Take `1 eV= 1.6 xx 10^(-19) J`
Answer» Correct Answer - A::B No of photons // sec `= (Energy incident on platimum surface per sec and )/(Energy of one photon )` No , of photons incident per second `= (2 xx 10 xx 10^(-4))/(10.6 xx 1.6 xx 10^(-19)) = 1.18 xx 10^(14)` As `0.53% ` of incident photon can eject photoelectron :. no of photoelectron eject per second `= 1.18 xx 10^(14) xx (0.53)/(100) = 6.25 xx 10^(11)` `minimum energy = 0eV` `Maximum energy = (10.6 - 5.6 = 5 eV`

Discussion

85.	The time period of the electron in the ground state of hydrogen atom is two times the times period of the electon in the first excited state of a certain hydrongen like atom (Atomic number Z). The value of Z isA. `2`B. `3`C. `4`D. None of these
Answer» Correct Answer - C `T = (2pir)/(v) or T prop (r )/(v) or T prop ((n^2//Z)/(Z//n))` `:. T prop (n)^3/(Z)^2` `((T_1)/(T_2)) = ((n_1)/(N_2))^3 = ((Z_2)/(Z_1))^2` `2 = ((1)/(2))^3 ((Z)/(1))^2` `:. Z= 4`

Discussion

86.	Light wave described by the equation `200 V//m sin (1.5xx10^15 s^(-1) t cos (0.5xx10^15 s^(-1) t` falls metal surface having work function 2.0 eV. Then, the maximum kinetic energy photoelectrons isA. `3.27eV`B. `2.2eV`C. `2.85eV`D. none of these
Answer» Correct Answer - D `omega_(max) = 1.5xx10^(15) rad/s = 2pif_(max)` `:. f_(max) = (1.5xx10)^15/(2pi) Hz` `E = (hf_(max)/(1.6xx10^(-19) eV` `=(6.63xx10^(-34)xx1.5xx10^(15)/(2pixx1.6xx10^(-19) eV` `~~1.0 eV` Since, E lt W, no photoemission can take place.

Discussion

87.	In a radioactive serries , `underset(92)overset(238)""` U change to `underset(82)overset(206)""` pb though `n_(1) alpha` -decay processses and `n_(2) beta-` decay processes.A. `n_(1)=8,n_(2)=8`B. `n_(1)=6,n_(2)=6`C. `n_(1)=8,n_(2)=6`D. `n_(1)=6,n_(2)=8`
Answer» Correct Answer - c

Discussion

88.	A photon of energy 10.2 eV corresponds to light of wavelength `lamda_(0)`. Due to an electron transition from n=2 to n=1 in a hydrogen atom, light of wavelength `lamda` is emitted. If we take into account the recoil of the atom when the photon is emitted.A. `lamda=lamda_(0)`B. `lamdaltlamda_(0)`C. `lamdagtlamda_(0)`D. the data is not suffcient to reach a cnclusion
Answer» Correct Answer - c

Discussion

89.	If a beam consisiting of `alpha,beta` and `gamma` radiation is passed through an electric field perpendicular to the beam, the deflections suffered by the components, in decreasing ordre are,A. `alpha,beta,gamma`B. `alpha,gamma,beta`C. `beta,alpha,gamma`D. `beta,gamma,alpha`
Answer» Correct Answer - c

Discussion

90.	If radiation of all wavelangths from ultavoilet to infraed is passed through hydrogen gas at room temperature , absorption lines will be observed in theA. Lyman seriesB. Balmer seriesC. both (a) and(b)D. neither (a) nor (b)
Answer» Correct Answer - a

Discussion

91.	When white light (violet to red ) is passed thouogh hydrogen gas at room temparature , absorption lines will be observed in theA. Lyman seriesB. Balmer seriesC. both (a) and(b)D. neither (a) nor (b)
Answer» Correct Answer - d

Discussion

92.	A fission reaction is given by `_(92)^(236) U rarr_(54)^(140) Xe + _(38)^(94)St + x + y` , where x and y are two particle Consider `_(92)^(236) U` to be at rest , the kinetic energies of the products are deneted by `k_(xe) K _(st) K _(s) (2MeV ) and ` repectively . Let the binding energy per nucleus of `_(92)^(236) U, _(54)^(140) Xe and _(38)^(94)St be 7.5 MeV , 8.4 MeV and 8.5 MeV, ` respectively Considering different conservation laws, the correct sption (s) is (are)A. `x = n , y= n , K_(St) = 129 MeV , K_(xe) = 86 MeV `B. `x = p , y= e^(bar), K_(St) = 129 MeV , K_(xe) = 86 MeV `C. `x = p, y = n , K_(St) = 129 MeV , K_(xe) = 86 MeV `D. `x = n , y= n , K_(St) = 86 MeV , K_(xe) = 129 MeV `
Answer» Correct Answer - A `_(92)^(236)U rarr _(54)^(140) Xe + _(38)^(94)St + xn + y ` The number of proton in reactants is equal to the products (lesving x and y ) and nujmber of product (leaving x and y ) is two less then reactants `:. X = p , y = e^(bar) ` is relied out [B] is increases and `x = p , y = n ` is rulled out [C] is increase `total energy less then = (236 xx 7.5 ) - [ 140 xx 8.5 + 94 xx 8.5] = 219 MeV` The energies of for and by together is `4 MeV` The energy remain is disctrthused by sr and Xe which is equal in `219 - 4 = 215 MeV` `:. `A is the currect option also momentum is conserved `:. K.E. prop (1)/(m) therefore K.E. _(m) gt K.E. _(we)` The energy of ke by togather is `4 MeV` The energy remain is distthuted by sr and Xe which is equal is `219 - 4 = 215 MeV` A is the correct option Also moomentum, is conserved Therefore `K.E. gt K.E. _(xe)`

Discussion

93.	Hydrogen `(_(1)H^(1))` Deuterium `(_(1)H^(2))` singly omised helium `(_(1)He^(1))` and doubly ionised lithium `(_(1)Li^(6))^(++)` all have one electron around the nucleus Consider an electron transition from `n = 2 to n = 1` if the wavelength of emitted radiartion are `lambda_(1),lambda_(2), lambda_(3),and lambda_(4)`, repectivelly then approximetely which one of the following is correct ?A. `4 lambda _(1) = 2 lambda _(2) = 2 lambda _(3)= lambda _(4)`B. `lambda _(1) = 2 lambda _(2) = 2 lambda _(3)= lambda _(4)`C. ` lambda _(1) = lambda _(2) = 4 lambda _(3)= 9 lambda _(4) `D. ` lambda _(1) = 2 lambda _(2) = 3 lambda _(3)= 4 lambda _(4) `
Answer» Correct Answer - C Wave number `(1)/(lambda) RZ^(2)[(1)/(n_(1)^(2)) -(1)/(n_(2)^(2))]` `rArr lambda prop (1)/(Z^(2))` By question `n = 1 and n_(1)= 2 ` Then `lambda_(1) = lambda_(2) = 4 lambda_(3) =9 lambda_(4) =

Discussion

94.	A particle of mass 3m at rest decays into two particles of masses m and 2m having non-zero velocities. The ratio of the de Broglie wavelengths of the particles `((lamda_1)/(lamda_2))` isA. `(1)/(2)`B. `(1)/(4)`C. 2D. None of these
Answer» Correct Answer - D

Discussion

95.	A particle of mass `M` at rest dacays into two particle of masses `m_(1)` and `m_(2) `having non zero velocity . The radio of the de Broglie wavelength . The ratio of the de Broglie wavelengtgh of the particle `lambda , _(1) // lambda_(2)` isA. `m_(1)//m_(2)`B. `m_(2)//m_(1)`C. `1.0`D. ` sqrt(m_(1))//sqrt(m_(2))`
Answer» Correct Answer - C Appling conservation of linear momentum, intial momentum = final momentum `0 = M_(1) NU_(1) - M_(2)NU_(2)rArr M_(1) NU_(1) M_(2)NU_(2)` = `Now , (lambda_(1))/(lambda_(2)) = (h// M_(1) NU_(1))/((h// M_(2) NU_(2)) = 1`

Discussion

96.	When an electron in the hydrogen atom in ground state absorb a photon of energy `12.1eV`, its angular momentumA. decreases by `2.11xx10^(-34) J-s`B. decreases by `1.055xx10^(-34) J-s`C. increase by `2.11xx10^(-34) J-s`D. increase by `1.055xx10^(-34) J-s`
Answer» Correct Answer - C

Discussion

97.	Let p and E denote the linear momentum and energy of a photon. If the wavelength is decreased,A. both p and E increaseB. p increase and E decreasesC. p decrease and E increasesD. both p and E decrreases
Answer» Correct Answer - A

Discussion

98.	The momentum of an x-ray photon with `lambda = 0.5 Å` isA. `13.26xx 10^(-26) kg- m//s`B. `1.326xx10^(-26) kg-m//g`C. `13.26xx10^(-24) kg-m//s`D. `13.26 xx10^(-22) kg-m//s`
Answer» Correct Answer - C `P = (h)/lambda)`

Discussion

99.	A photon has momentum of magnitude `8.24xx10^(-28) kg-m//s`. (a) What is the energy of this photon? Given your answer in joules and in electron volts (b) What is the wavelength of this photon? In what region of the electormagnetic spectrum does lie?
Answer» Correct Answer - A::B::D (a) `P = (E ) /(c ) rArr E = pc` (b) `lambda = (12375)/(E (in eV)` `=(12375)/(1.54) = 8035 Å` `~~804 nm` So, this wavelength lies in ultraviolet region.

Discussion

100.	Photon of frequency `v` has a momentum associated with it . If `c` is the velocity of light , the momentum isA. `hc//c`B. `v//c`C. `h v c `D. `h v//c^(2)`
Answer» Correct Answer - A Energy of a photom of frequency v is given by `E = hv ` also `E = mc^(2) , mc^(2)= hc` `rArr mc = (h v)/( c) rArr p = (h v)/( c)`

Discussion

Explore topic-wise InterviewSolutions in .

A sample of radioactive substance loses half of its activity in 4 days. The time in which its activity is reduced to 5% isA. (a) 12 daysB. (b) 8.3 daysC. (c) 17.3 daysD. (d) None of these

The activity of a radioactive sample goes down to about 6% in a time of 2 hour. The half-life of the sample in minute is aboutA. (a) 30B. (b) 15C. (c) 60D. (d) 120

A radioactive substance has a half-life of 64.8 h. A sample containing this isotope has an initial activity `(t=0)` of `40muCi`. Calculate the number of nuclei that decay in the time interval between `t_1=10.0h` and `t_2=12.0h`.

A radioactive sample contains `1.00xx10^15` atoms and has an activity of `6.00xx10^11` Bq. What is its half-life?

At time `t=0`, activity of a radioactive substance is 1600 Bq, at t=8 s activity remains 100 Bq. Find the activity at t=2 s.

At time `t=0` , number of nuclei of a radioactive substance are 100. At `t=1` s these numbers become 90. Find the number of nuclei at `t=2 s`.

In a certain nuclear reactor, a radioactive nucleus is bieng produced at a constant rate =1000/s The mean life of the radionuclide is 40 minutes. At steady state, the number of radionuclide will beA. `4xx10^(4)`B. `24xx10^(4)`C. `24xx10^(5)`D. `24xx10^(6)`

The combination of shown below yiedsA. `OR gate`B. `NOT gate`C. `XOR gate`D. `NAND gate`

A nucleus `._(Z)X^(A)` emits 2 `alpha -` particles and 3 `beta -` particles. The ratio of total protons and neutrons in the final nucleus is :A. `(Z - 7)/(A - Z + 7)`B. `(Z - 1)/(A - Z - 8)`C. `(Z - 1)/(A - Z - 7)`D. `(Z - 3)/(A - Z + 3)`

Stopping potential of 24, 100, 110 and 115 k V are measured for photoelectrons emitted from a certain element when it is radiated with monochromatic X-ray . If this element is used as a target in an X-ray tube,what will be the wavelength of `K_a` - line?

What will be the maximum kinetic energy of the photoelectrons ejected from magnesium (for which the work -function W=3.7 eV) when irradiated by ultraviolet light of frequency `1.5xx10^15 s^(-1)`.

The stopping potential for the photoelectrons emitted from a metal surface of work function 1.7 eV is 10.4 V. Find the wavelength of the radiaiton used. Also, identify the energy levels in hydrogen atom, which will emit this wavelength.

In a photeclectric experiment , when electromegnetic wave given by `E=E_(0)"sin"omegat` is incident , electron just ejects. When `E=E_(0)"sin"2omegat` is incident `K_("max")=K_(1)` and when E`=E_(0)"sin" 6omegat` is incident `K_("amx")=k_(2)`. When `k_(2)//k_(1)`

If a star can convert all the He nuclei completely into oxygen nuclei. The energy released per oxygen nuclei is (Mass of the helium nucleus is 4.0026 amu and mass of oxygen nucleus is 15.9994 amu)A. (a) 7.6 MeVB. (b) 56.12 MeVC. (c) 10.24 MeVD. (d) 23.4 MeV

At a specific instant emission of radioactive compound is deffernt in a field . The compound can emit (i) electron (ii)protons(iii)`He^(2+ )`(iv) neutrons The emission at instant can beA. I,ii,iiiB. I,ii,iii,ivC. ivD. ii,iii

An electron is excited from a lower energy state to a higher energy state in a hydrogen atom. Which of the following quantity/quantities decreases/decrease in the excitation ?A. potential energyB. Angular speedC. Kinetic energyD. Angularn momentum

An electron of kinetic enregy K collides elastically with a stationary hydrogen atom is the ground state. Then,A. `K gt 13.6 eV`B. `K gt 10.2 eV`C. `K lt 10.2 eV`D. data insufficient

In a hydrogen atom, the binding energy of the electron in the ground state is `E_(1)` then the frequency of revolution of the electron in the nth orbit isA. `2E_(1)//nh`B. `2E_(1)n//h`C. `E_(1)//nh`D. `E_(1)n//h`

Nuclei X and Y convert into a stable nucleus Z. At t=0 , the number of nuclei of X is 8 times that of Y. Half life of X is 1 hour and half life of y=2 hour. Find the time (in hour) at which rate of disintegration of X and Y are equal.

Regarding a nucleus choose the correct options.A. (a) Density of a nucleus is directly proportional to mass number AB. (b) Density of all the nuclei is almost constant of the order of `10^17kg//m^3`C. (c) Nucleus radius is of the order of `10^-15m`D. (d) Nucleus radius `prop` A

The atomic masses of the hydrogen isotopes are Hydrogen `m_1H^1=1.007825` amu Deuterium `m_1H^2=2.014102` amu Tritium `m_1H^3=3.016049` amu The mass of deuterium, `_1H^2` that would be needed to generate 1 kWhA. (a) `3.7kg`B. (b) `3.7g`C. (c) `3.7xx10^-5kg`D. (d) `3.7xx10^-8kg`

The atomic masses of the hydrogen isotopes are Hydrogen `m_1H^1=1.007825` amu Deuterium `m_1H^2=2.014102` amu Tritium `m_1H^3=3.016049` amu The number of fusion reactions required to generate 1kWh is nearlyA. (a) `10^8`B. (b) `10^18`C. (c) `10^28`D. (d) `10^38`

The atomic masses of the hydrogen isotopes are Hydrogen `m_1H^1=1.007825` amu Deuterium `m_1H^2=2.014102` amu Tritium `m_1H^3=3.016049` amu The energy released in the reaction, `_1H^2+_1H^2rarr_1H^3+_1H^1` is nearlyA. (a) 1MeVB. (b) 2MeVC. (c) 4MeVD. (d) 8 MeV

The time period of the electron in the ground state of hydrogen atom is two times the times period of the electon in the first excited state of a certain hydrongen like atom (Atomic number Z). The value of Z isA. `2`B. `3`C. `4`D. None of these

Light wave described by the equation `200 V//m sin (1.5xx10^15 s^(-1) t cos (0.5xx10^15 s^(-1) t` falls metal surface having work function 2.0 eV. Then, the maximum kinetic energy photoelectrons isA. `3.27eV`B. `2.2eV`C. `2.85eV`D. none of these

In a radioactive serries , `underset(92)overset(238)""` U change to `underset(82)overset(206)""` pb though `n_(1) alpha` -decay processses and `n_(2) beta-` decay processes.A. `n_(1)=8,n_(2)=8`B. `n_(1)=6,n_(2)=6`C. `n_(1)=8,n_(2)=6`D. `n_(1)=6,n_(2)=8`

If a beam consisiting of `alpha,beta` and `gamma` radiation is passed through an electric field perpendicular to the beam, the deflections suffered by the components, in decreasing ordre are,A. `alpha,beta,gamma`B. `alpha,gamma,beta`C. `beta,alpha,gamma`D. `beta,gamma,alpha`

If radiation of all wavelangths from ultavoilet to infraed is passed through hydrogen gas at room temperature , absorption lines will be observed in theA. Lyman seriesB. Balmer seriesC. both (a) and(b)D. neither (a) nor (b)

When white light (violet to red ) is passed thouogh hydrogen gas at room temparature , absorption lines will be observed in theA. Lyman seriesB. Balmer seriesC. both (a) and(b)D. neither (a) nor (b)

A particle of mass 3m at rest decays into two particles of masses m and 2m having non-zero velocities. The ratio of the de Broglie wavelengths of the particles `((lamda_1)/(lamda_2))` isA. `(1)/(2)`B. `(1)/(4)`C. 2D. None of these

When an electron in the hydrogen atom in ground state absorb a photon of energy `12.1eV`, its angular momentumA. decreases by `2.11xx10^(-34) J-s`B. decreases by `1.055xx10^(-34) J-s`C. increase by `2.11xx10^(-34) J-s`D. increase by `1.055xx10^(-34) J-s`

Let p and E denote the linear momentum and energy of a photon. If the wavelength is decreased,A. both p and E increaseB. p increase and E decreasesC. p decrease and E increasesD. both p and E decrreases

The momentum of an x-ray photon with `lambda = 0.5 Å` isA. `13.26xx 10^(-26) kg- m//s`B. `1.326xx10^(-26) kg-m//g`C. `13.26xx10^(-24) kg-m//s`D. `13.26 xx10^(-22) kg-m//s`

A photon has momentum of magnitude `8.24xx10^(-28) kg-m//s`. (a) What is the energy of this photon? Given your answer in joules and in electron volts (b) What is the wavelength of this photon? In what region of the electormagnetic spectrum does lie?

Photon of frequency `v` has a momentum associated with it . If `c` is the velocity of light , the momentum isA. `hc//c`B. `v//c`C. `h v c `D. `h v//c^(2)`