InterviewSolution
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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.
| 1. |
Statement I : Threshold frequency supports wave nature of light. Statement II : Photo electrons are not emitted when incident light has frequency lesser than threshold frequency.A. If both Statement- I and Statement- II are true, and Statement - II is the correct explanation of Statement– I.B. If both Statement - I and Statement - II are true but Statement - II is not the correct explanation of Statement – I.C. If Statement - I is true but Statement - II is false.D. If Statement - I is false but Statement - II is true. |
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Answer» Correct Answer - D |
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| 2. |
A radioactive substance is being consumed at a constant of `1 s^(-1)`. After what time will the number of radioactive nuclei becoem `100`. Initially, there were 200 nuceli present.A. 1 sB. `(1)/(ln(2))s`C. `ln (2) s`D. 2 s |
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Answer» Correct Answer - C |
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| 3. |
Assuming the splitting of `U^235` nucleus liberates 200 MeV energy, find (a) the energy liberated in the fission of 1 kg of `U^235` and (b) the mass of the coal with calorific value of `30 kJ//g` which is equivalent to 1 kg of `U^235`. |
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Answer» Correct Answer - A::B::C (a) Number of nuclei in 1 kg of `U^235`, `N=(1/235)(6.02xx10^26)` `:.` Total energy released `=(Nxx200)MeV` `=(1/235)(6.02xx10^26)(200)(1.6xx10^-13)` `=8.19xx10^13J` (b) `m=(8.19xx10^13)/(30xx10^3)g` `=2.73xx10^9g` `=2.73xx10^6kg` |
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| 4. |
The energy released by the fission of a single uranium nucleus is 200 MeV. The number of fission of uranium nucleus per second required to produce 16 MW of power is (Assume efficiency of the reactor is 50%)A. (a) `2xx10^6`B. (b) `2.5xx10^6`C. (c) `5xx10^6`D. (d) None of these |
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Answer» Correct Answer - D It means we are getting only 100 MeV of energy by the fission of one uranium nucleus. Number of nuclei per second =Energy required per second/Energy obtained by one fission `=(16xx10^6)/(100xx1.6xx10^-13)=10^18` |
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| 5. |
A radioactive isotope is being produced at a constant rate `dN//dt=R` in an experiment. The isotope has a half-life `t_(1//2)`.Show that after a time `t gtgt t_(1//2)`,the number of active nuclei will become constant. Find the value of this constant.A. (a) ATB. (b) `A/T1n2`C. (c) AT 1n2D. (d) `(AT)/(1n2)` |
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Answer» Correct Answer - D When the rate producion = rate of disintegration, number of nuclei or maximum. `:.` `lambdaN=A` or `(1n2)/TN=A` or `N=(AT)/(1n2)=`maximum |
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| 6. |
A radiaocatice isotope is being produced at a constant rate `X`. Half-life of the radioactive substance is `Y`. After some time, the number of radioactive nuceli become constant. The value of this constant is .A. `(XY)/(ln(2))`B. XYC. (XY) ln (2)D. `(X)/(Y)` |
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Answer» Correct Answer - A |
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| 7. |
A radioactive element` X` converts into another stable elemnet `Y`. Half-life of `X` is `2h`. Initally, only `X` is present. After time `t`, the ratio of atoms of `X` and `Y` is found to be `1:4` Then `t` in hours is .A. 2B. 4C. between 4 and 6D. 4 |
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Answer» Correct Answer - C |
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| 8. |
A nucleus X, initially at rest , undergoes alpha dacay according to the equation , ` _(92)^(A) X rarr _(2)^(228)Y + a ` (a) Find the value of `A` and `Z` in the above process. (b) The alpha particle producted in the above process is found to move in a circular track of radius `0.11 m` in a uniform mmagnatic field of `3` Tesia find the energy (in MeV) released during the process and the binding energy of the patent nucleus X Given that `: m (gamma) = 228.03 u, m(_(0)^(1)) = 1.0029 u. ` `m (_(2)^(4) He) = 4.003 u , m (_(1)^(1) H) = 1.008 u ` |
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Answer» Correct Answer - A::B::C::D (a)`_(92)^(A) X rarr _(x)^(228) Y + _(2)^(4)He` `A = 228 + 4 = 232.92 = z = 2 rArr Z = 90` (b) Let v be the velocity with which a particle is emitted Then ` (m nu^(3))/(r ) = q nu B rArr nu = (q r B)/(m) = (2 xx 1.6 xx 10^(-19) xx 0.11 xx 3)/(4.003 xx 10^(-27))` `rArr nu = 1.59 xx 10^(7) ms^(-1)` Appling law of conservation of linear momentum during a - decay we get `m_(1) nu_(1) = m_(0)nu_(0)` ....(i) The total kinetic energy of a particle and `Y` is `E = K.E_(0) = + K.E_(y) = (1)/(2) m_(0) nu_(0)^(2) + (1)/(2) m_(1)nu_(1)^(2)` ` = (1)/(2)m_(0) nu_(0)^(2) + (1)/(2) m nu [(m_(0)nu_(0))/(m_(y))]^(2) = (1)/(2) m_(0)nu_(0)^(2) + m+(0)^(2) + (m_(0)^(2) nu _(0)^(2))/(2 m y)` `= (1)/(2) m_(0) nu_(0)^(2) [1 +(m_(0))/(m nu)]` `= (1)/(2) xx 4.003 xx 1.6 xx 10^(-27) xx (1.59 xx 10^(2)) ^(2) [ 1 + (4.003)/(228.03)]J` `= 8.55 xx 10^(-13) J` `= 5.34 MeV` :. ` Mass equivalent of this energy ` = (5.34)/(931.5) = 0.0051 a m u ` Also `m_(2) = m_(y) + m_(a) + ` mass equivalent of energy (E) `= 228.03 + 4.003 = 0.0057 = 232.0387 a` The number of nucleus `= 92 protons + 140 nutroun ` `:.` Bimdimng energy of nucleus X `= (92 xx 1.006 + 140 xx 1.009) - 232.0387 = 1.971 u` `= = 1.9971 xx 931.5 = 1023 eN` |
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| 9. |
The half - life of `^(215)` At is `100 mu , s.`The time taken for the radioactivity of a sample of `^(215)` At to dacay to `1//16^(th)` of its initialy value isA. `400 mu s `B. `63 mu s `C. `40 mu s `D. `300 mu s ` |
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Answer» Correct Answer - A `A = A_(0) (1//2)^(n) , n= `number of half lives . `(A_(0))/(16) =A_(0) ((1)/(2)) ^(n) :. ((1)/(2)) ^(4) = ((1)/(2)) ^(n) ` `rArr n = 4` `:. T = (4 xx 100)mus = 400 mus` |
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| 10. |
X-rays are produced by impinging_______on a target.A. `alpha` particlesB. protonsC. electronsD. neutrons |
| Answer» Correct Answer - C | |
| 11. |
A modified discharge tube is used as ________.A. cathode ray oscilloscopeB. fluorescent tubeC. X-ray tubeD. All the above |
| Answer» Correct Answer - D | |
| 12. |
The discharge tube is filled with uniform_______column glow at how enough pressure of the gas in the tube. |
| Answer» Correct Answer - luminous or positive | |
| 13. |
The conditions for the discharge of electricity through gases in a discharge tube is _______.A. high potential and high pressureB. high potential and low pressureC. low potential and high pressureD. low potential and low pressure |
| Answer» Correct Answer - B | |
| 14. |
In a Coolidge tube, an electric field is appliedA. to increases the charge on the cathode rays.B. to accelerate the cathode rays.C. to produce fluorescenceD. to decrease the charge on the cathode rays. |
| Answer» Correct Answer - B | |
| 15. |
The false statement about X-rays is________.A. X-rays are not particlesB. X-rays are unchargedC. X-rays can penetrate through all bodiesD. X-rays cause fluorescence when they are incident on cadmium sulphide |
| Answer» Correct Answer - C | |
| 16. |
The fluorescence of the glass (discharge) tube at very low pressure is characteristic of_________.A. the phosphors in the material of the glassB. the gas used in the tubeC. the cathodeD. the emitted particles |
| Answer» Correct Answer - A::B::C::D | |
| 17. |
Two radioactive `X_(1)` and `X_(2)` have decay constants `10 lambda ` and `lambda` respectively . If initially they have the same number of nuclei, then the ratio of the number of nuclei of `X_(1)` to that of `X_(2)`will be `1//e` after a time . |
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Answer» Correct Answer - A::B::D `(N_(x_1)(t))/(N_(x_2)(t))=1/e` or `(N_0e^(-10lambdat))/(N_0e^(-lambdat)) = 1/e` (Initially, both have same number of nuclei say N `_(0)`) or `e^(-9lambdat)=e^-1` or `e=e^(9lambdat)` or `9lambdat = 1` or `t=1/(9lambda)` :. The correct option is (d). |
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| 18. |
A nucleus X initially at rest, undergoes alpha decay according to the equation `_Z^232Xrarr_90^AY+alpha` What fraction of the total energy released in the decay will be the kinetic energy of the alpha particle?A. (a) `90/92`B. (b) `228/232`C. (c) `sqrt(228/232)`D. (d) `1/2` |
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Answer» Correct Answer - B `A=232-4=228` From conservation of momentum, `p_alpha=p_gamma=sqrt(2K_alpham_alpha)=sqrt(2K_gammam_gamma)` or `(K_alpha)/(K_gamma)=(m_gamma)/(m_alpha)=228/4` `:.` `K_alpha=((228)/(228+4))K_(t otal)` `=(228/232)K_(T otal)` |
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| 19. |
Half-life of a radioactive substance is T. At time t_1 activity of a radioactive substance is `R_1` and at time `t_2` it is `R_2`. Find the number of nuclei decayed in this interval of time. |
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Answer» Correct Answer - A::B Half-life is given by `t_(1//2) = (1n2)/(lambda)` :. `lambda = (1n2)/(t_(1//2))=1n2/T` Activity `R= lambdaN` :. `N=R/lambda=(RT)/(1n2)` When activity is `R_1`, numbers of nuclei are `N_1 =(R_1T)/(1n2)` Similarly, `N_2 = (R_2T)/(1n2)` :. Numbers decayed `=N_1-N_2 = ((R_1-R_2)T)/(1n2)` |
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| 20. |
An unstable nucleus X can decay into two stable nuclie Y and Z A sample containing only X is taken at t=0 Three graphs `log_(e)(N_(x)` vs t,`N_(x),` vs t and `N_(z)` vs t are drawn as shown below Here `N_(x),N_(y)` and `N_(z)` represent number of nuclei of X,Y and Z repectively at any instant. t Choose the correct choice (s) from the following :A. Decay constant for decay of X into Y is `(b tan theta)/(e^(a))`B. ecay constant for decay of X into Z is `(c tan theta)/(e^(a))`C. Number of nuclei of X at t=0-is `e^(a)`D. Half-life of nuclei X is `(1)/(tan theta)` |
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Answer» Correct Answer - A::B::C |
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| 21. |
Two radioactive samples of different elements (half-lives `t_1` and `t_2` respectively) have same number of nuclei at `t=0`. The time after which their activities are same isA. (a) `(t_1t_2)/(0.693(t_2-t_1))1n(t_2)/(t_1)`B. (b) `(t_1t_2)/(0.693)1n(t_2)/(t_1)`C. (c) `(t_1t_2)/(0.693(t_1+t_2))1n(t_2)/(t_1)`D. (d) None of these |
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Answer» Correct Answer - A `R_1=R_2` `R_(01)e^(-lambda_1t)=R_(02)e^(-lambdat)` `:.` `lambda_1N_0e^(-lambda_1t)=lambda_2N_0e^(-lambda_2t)` …(i) `lambda_1N_0e^(-lambda_1t)=lambda_2N_0e^(-lambda_2t)` …(i) `lambda_1=(1n2)/(t_1)=0.693/t_1` `lambda_2=(0.693)/(t_2)` Substituting these values in Eq. (i), we can get the t. |
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| 22. |
Assertion : Only those nuclei which are heavier than lead are radioactive. Reason : Nuclei of elements heavier than lead are unstable.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 Some lighter nuclei are also radioactive. |
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| 23. |
The atom of a heavy fissionable element hit by a neutron of sufficient energy breaks into two or more lighter elements with the release of two or additional neutrons because-A. Neutron is an uncharged particleB. Momentum of neutron is very largeC. It is easier for protons than neutrons to be in the nucleusD. Neutron-proton ratio increases as mass number of the element increases |
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Answer» Correct Answer - A |
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| 24. |
Assertion: The nuclear energy can be obtained by the nuclear fission of heavier nuclei as well as by fusion of lighter nuclei. Reason: As the mass number increases, the binding energy per nucleon, first increases and then decreases.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 - A::B In binding energy per nucleon versus mass number graph binding energy per nucleon of daughter nuclei should increase (for release of energy) or the daughter nuclei should lie towards the peak of the graph. |
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| 25. |
When the nucleus o fan eletrically neutral atom undergoes a radioactive decay process , it will remain after he decay if the process isA. an`alpha` decayB. a`beta^(-)` decayC. a `gamma` decayD. a k-capture process |
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Answer» Correct Answer - c,d |
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| 26. |
Two identical samples (same material and same amout) P and Q of a radioactive substance having mean life T are observed to have activities `A_P` and `A_Q` respectively at the time of observation. If P is older than Q, then the difference in their age isA. (a) `T 1n ((A_P)/(A_Q))`B. (b) `T 1n ((A_Q)/(A_P))`C. (c) `T((A_P)/(A_Q))`D. (d) `T((A_Q)/(A_P))` |
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Answer» Correct Answer - B `A_P=A_0e^(-lambdat_1)` `A_Q=A_0e^(-lambdat_2)` `lambda_(t_1)=1n(A_0//A_P)` `:.` `t_1=1/lambda1n(A_0//A_P)=T1n(A_0//A_P)` Similarly, `t_2=T1n(A_0//A_Q)` `:.` `t_1-t_2=T1n((A_Q)/(A_P))` |
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| 27. |
A mixture of hydrogen atoms (in their ground state) and hydrogen like ions (in their first excited state) are being excited by electrons which have been accelerated by same potential of emitted light are found in the ratio 5:1. Then, find (a) the minimum value of V for which both the atoms get excited after collision with electrons. (b) atomic number of other ion. (c ) the energy of emitted light. |
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Answer» Correct Answer - A::B::C::D (a) and (b) when hydrogen atom is excited, then `eV = E_0 ((1)/(1) - (1)/(n^2)) ….(i)` When ion is excited, `eV = E_0Z^2[(1)/(2^2) - (1)/(n_1^2)] …..(ii)` Wavelength of emitted light, `(hc)/(lambda_1) = E_0((1)/(1) - (1)/(n^2)) ......(iii)` `(hc)/(lambda_2) = E_0Z^2 ((1)/(1) - (1)/(n_1^2)) .....(iv)` Further it is given that `(lambda_1)/(lambda_2) = (5)/(1) ....(v)` Solving the above equations, we get `Z =2, n=2, n_1 =4` and V = 10.2 V (c) Energy of emitted photon by the hydrogen atom `= E_2 - E_1` = 10.2 eV and by the ion `= E_4 - E_1` `=(13.6)(2)^2 ((1-(1)/(16))` = 51eV. |
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| 28. |
An electron is orbiting is a circular orbit of radius r under the influence of a constant magnetic field of strength B. Assuming that Bohr postulate regarding the quantisation of angular momentum holds good for this elerctron, find (a) the allowed values of te radius r of the orbit. (b) the kinetic energy of the electron in orbit (c ) the potential energy of insteraction between the magnetic moment of the orbital current due to the electron moving in its orbit and the magnetic field B. (d) the total energy of the allowed energy levels. (e) the total magnetic flux due to the magnetic field B passing through the nth orbit. (Assume that the charge on the electronis -e and the mass of the electron is m). |
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Answer» Correct Answer - B::C::D (a) `r = (mv)/(Be) ….(i)` `mvr = (nh)/(2pi) ….(ii)` Solving these two equations, we get `r = sqrt((nh)/(2piBe))` and` v = sqrt((nhBe)/(2pim^2))` (b) `K = (1)/(2) mv^2 = (nhBe)/(4pim)` (c ) `M = iA =((e )/(T)) (pir^2)` `=(e )/((2pir)/(v)) (pir^2) = (evr)/(2)`` `=` (e )/(2) sqrt((nh)/(2piBe)) sqrt((nhBe)/(2pim^2)` `=(nhe)/(4pim)` `U =- MB cos 180^@` `=(nhe)/(4pim).` |
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| 29. |
Ne nucleus , the after absorbing energy , decays into two `a - particle` and an unknown nucleus . The unknown nucleus isA. nitrogenB. carbonC. boronD. oxygen |
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Answer» Correct Answer - B `_(10)^(22)Ne rarr _(2)^(4)He + _(2)^(4)He + _(6)^(14) X ` The new element `X` has atomic number `6` . Therefore , it is carbon atom . |
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| 30. |
The ground state and first excited state energies of hydrogen atom are `-13.6 eV` and `-3.4eV` respectively. If potential energy in ground state is taken to be zero. Then:A. potential energy in the first excited state would be `20.4 ev `B. total energy in the first excited state would be `23.8 eV`C. kinetic energy in the first excited state would be `3.4 eV`D. total energy in the ground state would be `13. 6 eV` |
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Answer» Correct Answer - A::B::C::D |
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| 31. |
A hydrogen atom in its ground state is by light of wavelength ` 970Å Taking hc//e = 1.237 xx 10^(-6)eV m` and the ground state energy of hydrogen atom as ` - 13.6 eV` the number of lines present in the emmission spectrum is |
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Answer» `E = (hc)/(lambda) = (1.237 xx 10^(-6))/(970 xx 10^(-10)) eV = 12.75 eV` `:. `The energy of electron after absorbing this photon `= - 13.6 + 12.75 = - 0.85eV ` This corresponds to `n = 4 ` Number of spectral line `= (n(n-1))/(2) = (4(4 - 1))/(2) = 6 ` |
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| 32. |
Binding energy per nucleon for `C^(12)` is `7.68 MeV` and for `C^(13)` is `7.74 MeV`. The energy required to remove a neutron from `C^(13)` is .A. `5.34 MeV`B. `5.5 MeV`C. `9.5 MeV`D. `9.34 MeV` |
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Answer» Correct Answer - A |
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| 33. |
If the binding energy per nucleon in `_(3)^(7) Li and _(2)^(4)He nuclei are 5.60 MeV and 7.06MeV` respectively then in the reaction `P +_(3)^(7) Li rarr 2 _(2)^(4) He` energy of proton mnust beA. `28.24 MeV`B. `17.28 MeV`C. `1.46 MeV`D. `39.2 MeV` |
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Answer» Correct Answer - B Let `E` be the energy of proton , than `E + 7 xx 5.6 = 2 xx [4 xx 7.06]` `rArr E = 56.48 - 39.2 = 17.28MeV` |
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| 34. |
If the following which one of the diodes reverse biased?A. B. C. D. |
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Answer» Correct Answer - D p- side connected to low potential and n- side is connected to high potential |
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| 35. |
The anode vollage of a photocell is kept fixed . The wavelength `lambda `of the light falling on the cathode varies as followsA. B. C. D. |
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Answer» Correct Answer - B As `lambda` decreases .v increases and hence the speed of photoelectron increases . The chance of photoelectron to meet the anod increases |
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| 36. |
The anode vollage of a photocell is kept fixed . The wavelength `lambda `of the light falling on the cathode varies as followsA. B. C. D. |
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Answer» Correct Answer - D As `lambda` is increased there will be a value of `3` above which photoelectrons will be cases to come out so photocurrect will became zero heance (d) is correctanswer |
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| 37. |
In a mixture of `H- He^(+)` gas (`He+ `is singly ionized He atom), `H` atom and `He+` ions are excited to their respective first sxcited state subequendy `H` atoms transfer their total excilation energy to `He+` ions (by collsions) Assume that the bohr model of atom is exactly veld. The questum number `n` of the state fnally populand in `He^(+)` inos is -A. `2`B. `3`C. `4`D. `5` |
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Answer» Correct Answer - C For hydrogen like atom `E_(0) = (-13.6 Z^(2))/(n^(2))eV // atom ` `(for hydrogebn atom)/((Z= 1))` E_(1) = - 13.6 eV` `E_(2) = -3.4 eV` :.delta E = E_(2) - E_(1) = - 3.4 - (-13.6 ) = 10.2 eV` i.e. when hydrogen comes to ground state it will release `10.2 eV` of energy `(for He^(+)ion)/((Z = 2))` E_(1) = - 13.6 xx 4 eV = - 54.4 eV ` `E_(2) = -13.6eV` `E_(3) = - 6.04eV` `E_(4) = - 3.4 eV` Here `He^(+)` ion is in first excited state i. e. possessing energuy `= 13.6eV` . After receiving of `+ 10.2 eV` from excited hydrogen atom on collsion , the energy of electron will be `(-13.6 + 10.2) eV = -3.4 eV` This means that the electron will jump to `n = 4` |
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| 38. |
Highly energetic electron are bombarded in a target of an element containing `30` neutrons Tne ratio of nucleus to that of Helium nucleus is `(14)^(1//3)` . Find (a) atomic number of the nucleus (b) the frequency of `k_(a)` line of the X- rays producted `(R = 1.1 xx 10^(7) m^(-1) and c = 3 xx 10^(8)m//s)` |
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Answer» Correct Answer - A::B::D KEY CONCEPT : we know that radius of nucleus is given by formula `r = r_(0)A^(1//3) where r_(0) constt , and A = mass number ` `for the nucleus r_(1) = r_(0) 4^(1//3)` for unknown nucleus `r_(2) = r_(0)4^(1//3)` `:. (r_(2))/(r_(1)) = ((d)/(4))^(1//3) , (14)^(1//3)= ((A)/(4))^(1//3) rArr A = 56` `:. `no of photon `= A - no of neucleus = 56 - 30 = 26` `:. atomic number = 26 ` we know that ` v = Rc (Z - b)^(2)[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` `here R = 1.1 xx 10^(7), c = 3 xx 10^(8) , Z = 26 ` ` h = 1 (for k_(a)) , n_(1) = 1 , n_(2) = 2 ` `:. n = 1.1 xx 10^(7) xx 3 xx 10^(8) (26 - 1 )^(2) [(1)/(1) - (1)/(4)]` `= 3.3 xx 10^(15) xx 25 xx 25 xx (3)/(4) = 1.546 xx 10^(16) Hz` |
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| 39. |
A radioactive nucleus X decays to a nucleus Y with a decay constant `lambda_X=0.1s^-1`, Y further decays to a stable nucleus Z with a decay constant `lambda_Y=1//30s^-1`. Initially, there are only X nuclei and their number is `N _0=10^20`. Set up the rate equations for the populations of X, Y and Z. The population of Y nucleus as a function of time is given by `N_Y(t)={N _0lambda_X//(lambda_X-lambda_Y)}[exp(-lambda_Yt)-exp(-lambda_Xt)]`. Find the time at which `N_Y` is maximum and determine the population X and Z at that instant. |
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Answer» Correct Answer - A::D (a) Let at time `t=t`, number of nuclei of Y and Z are `N_Y` and `N_Z`. Then, Rate equations of the populations of X, Y and Z are `((dN_X)/(dt))=-lambda_XN_X` …(i) `((dN_Y)/(dt))=lambda_XN_X-lambda_YN_Y`…(ii) and `((dN_Z)/(dt))=lambda_YN_Y`...(iii) (b) Given, `N_Y(t)=(N_0lambda_X)/(lambda_X-lambda_Y)[e^(-lambda_Yt)-e^(-lambda_Xt)]` For `N_Y` to be maximum `(dN_Y(t))/(dt)=0` i.e `lambda_XN_X=lambda_YN_Y` ...(iv) [from Eq. (ii)] or `lambda_X(N_0e^(-lambda_Xt))=lambda_Y(N_0lambda_X)/(lambda_X-lambda_Y)[e^(-lambda_Yt)-e^(-lambda_Xt)]` or `(lambda_X-lambda_Y)/(lambda_Y)=(e^(-lambda_Yt))/(e^(-lambda_Xt))-1` `lambda_X/lambda_Y=e^((lambda_X-lambda_Y)t)` or `(lambda_X-lambda_Y)t1n(e)=1n(lambda_X/lambda_Y)` or `t=(1)/(lambda_X-lambda_Y)1n(lambda_X/lambda_Y)` Substituting the values of `lambda_X` and `lambda_Y`, we have `t=(1)/((0.1-1//30))1n((0.1)/(1//30))=151n(3)` or `t=16.48s` (c) The population of X at this moment, `N_X=N_0e^(-lambda_Xt)=(10^20)e^(-(0.1)(16.48))` `N_X=1.92xx10^19` `N_Y=(N_Xlambda_X)/(lambda_Y)` [From Eq. (iv)] `=(1.92 xx 10^19)((0.1)/(1//30))` `=5.76xx10^19` `N_Z=N_0-N_X-N_Y` `=10^20-1.92xx10^19-5.76xx10^19` `N_Z=2.32xx10^19` |
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| 40. |
A radioactive element decays by `beta-emission`. A detector records n beta particles in 2 s and in next 2 s it records 0.75 n beta particles. Find mean life correct to nearest whole number. Given ln `|2| = 0.6931`, ln `|3|=1.0986`. |
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Answer» Let `n_0` be the number of radioactive nuclei at time `t=0`. Number of nuclei decayed in time t are given by `n_0(1-e^(-lambdat))`, which is also equal to the number of beta particles emitted during the same interval of time. For the given condition, `n=n_0(1-e^(-2lambda))` ...(i) `(n+0.75n)=n_0(1-e^(-4lambda))` ...(ii) Dividing Eq. (ii) by Eq. (i), we get `1.75=(1-e^(-4lambda))/(1-e^(-2lambda))` or `1.75 - 1.75 e^(-2lambda) = 1-e^(-4lambda)` :. `1.75e^(-2lambda)-e^(-4lambda)=3/4` ...(iii) Let us take `e^(-2lambda)=x` Then, the above equation is `x^2-1.75x+0.75=0` or `x=(1.75+-sqrt((1.75)^2-(4)(0.75)))/(2)` or `x=1 and 3/4` :. From Eq. (iii) either `e^(-2lambda)=1` or `e^(-2lambda)=3/4` but `e^(-2lambda)=1` is not acceptable because which means `lambda=0`. Hence, `e^(-2lambda)=3/4` or `-2lambda1n(e)=1n(3)-1n(4)=1n(3)-21n(2)` :. `lambda=1n(2)-(1)/(2)1n(3)` `Substituting the given values, `lambda=0.6931-1/2xx(1.0986)=0.14395s^-1` :. Mean life, `t_(mean)=1/lambda=6.947`s :. The correct answer is 7. |
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| 41. |
As a given instant there are `25%` undercayed radio - active nucles in az sample . After `10` second the number of undecayed nucles reduces to `12.5%` Calculate (i() mean - like of the nucleus, and (ii) the time in which the number of undecayed nuclei will further to `6.25 %` of the reducted number . |
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Answer» Correct Answer - A::C::D From the given in information , it is clear that half life of the radioactive nuclie is `10 sec` (since half the amount is consumed in `10 sec 12.5% ` is half of `25 %` pis note) Mean life ` tau = (1)/(lambda) = (1)/(0.693//t_(1//2)) = (t_(1//2))/(0.693) = (10)/(0.693) = 14.43 sec ` (ii) N = N_(0) e ^(- lambda t) rArr (N)/(N_(0)) = (6.25)/(100)` `lambda = 0.693 s^(-1)` `(6.25)/(100) = e^(-0.0693t) rArr e^(-0.0693t) = (100)/(6.25) = 16` `0.0693 t = in 16 = 2.773 or t =(2.733)/(0.0693) = 40 sec` |
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| 42. |
At a given instant there are 25% undecayed radioactive nuclei in a sample. After 10 s the number of undecayed nuclei reduces to 12.5%. Calculate (a) mean life of the nuclei, (b) the time in which the number of undecayed nuclei will further reduce to 6.25% of the reduced number. |
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Answer» Correct Answer - A::C::D (a) In 10 s, number of nuclei has been reduced to half (25% to 12.5%). Therefore, its half-life is `t_(1//2) = 10`s Relation between half-life and mean life is `t_(mean)=(t_(1//2))/(1n2)=10/0.693s` `t_(m ean)=14.43s` (b) From initial 100% to reduction till 6.25%, it takes four half-lives. `100%overset(t_(1//2))rarr50%overset(t_(1//2))rarr25%overset(t_(1//2))rarr12.5%overset(t_(1//2))rarr6.25%` `t=4t_(1//2)=4(10)s=40s` `t=40s` |
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| 43. |
The equation `4[H^(+) rarr _(2)^(4) He^(2+) + 2e bar + 26 MeV represents`A. ``beta - decay`B. `lambda - decay`C. `flasion`D. `fission` |
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Answer» Correct Answer - C `4 _(1)^(1) H^(+) rarr _(2)^(4)He^(2+) + 2 e ^(bar) + 26 MeV` represent a funtion reaction . |
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| 44. |
In Bohr model of the hydrogen atom, let R,v and E represent the radius of the orbit, speed of the electron and the total energy respectively. Which of the following quantities are directly proportional to the quantum number n?A. VRB. REC. `(v)/(E )`D. `(R )/(E )` |
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Answer» Correct Answer - a,c |
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| 45. |
A freshly prepared smaple contains `16xx10^(20)` raadioactive nuclei, whose mean life is `10^(10)` seconds The acitiivity of the sample just after 4 ahlr lives time isA. `(ln2)/(16)xx10^(10)dps`B. `1xx10^(10) dps`C. `(1)/(ln2)xx10^(10) dps`D. `ln 2xx10^(10) dps` |
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Answer» Correct Answer - B |
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| 46. |
Assertion: In continuous X-ray spectrum, all wavelength can be otained. Reason: Accelerated (or retarded) charged particles rediate enrgy. This is the cause of production of continuous X-rays.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason or true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D In X-ray spectrum, all wavelengths greater than `lambda_(min)` are obtained. |
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| 47. |
Carbon , silicon and germanium have four velence electrons each . At room temperature which one of the following statements is most appropriate ?A. the number of free electron for conduction is significant is `Si and Ge` but small in `C`B. the number of free conduction electron is significant in `C` but small in `Si and Ge`C. the number of free conduction electron is negligibly small in all the three.D. the number of free electron for conduction is significant in allthe three |
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Answer» Correct Answer - A `Si and Ge `are semiconductors but `C` is an insulator Also , the conduactively of `Si and Ge` is more then `C` because the velence electron of `Si,Ge and C ` lie in third fourth and second orbit respectively |
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| 48. |
A radioactive element disintegrates for a time interval equal to its mean life. The fraction that has disintegrated isA. (a) `1/e`B. (b) `1-1/e`C. (c) `0.693/e`D. (d) `0.693(1-1/e)` |
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Answer» Correct Answer - B Number of atoms disintegrated, `N=N_0(1-e^(-lambdat))` `:.` `N/N_0=1-e^(-lambdat)` At `t=`one mean life `=1/lambda` `N/N_0=1-e^-1=1-1/e` |
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| 49. |
Starting with a sample of pure `.^66Cu`, `3/4` of it decays into `Zn` in `15` minutes. The corresponding half-life isA. `15 minutes `B. `10 minutes`C. `7 (1)/(2) minutes `D. `5 minutes` |
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Answer» Correct Answer - D `(7)/(8) `of `Cu` decay in `15minutes` `:. `Cu undecay = N = 1 - (7)/(8) = (1)/(8) = ((1)/(2))^(3)` `:. ` no of half lifes `= 3` `n = (I)/(T) or 3 = (15)/(T)` `rArr T = half life period = (15)/(3) = 5 minutes` ` |
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| 50. |
Starting with a sample of pure `^66Cu`, `3/4` of it decays into Zn in 15 minutes. The corresponding half-life isA. (a) 5 minutesB. (b) 7.5 minutesC. (c) 10 minutesD. (d) 3.5 minutes |
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Answer» Correct Answer - B Decayed fraction is 3/4th. Therefore, left fraction is 1/4th. `N=N_0((1)/(2))^n` or `N/N_0=((1)/(2))^n=1/4` `:.` `n=` number of half-lives=2 Two half-lives are equivalent to 15 min. Therefore, one half-life is 7.5 min. |
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