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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.
| 301. |
A `""^118Cd` radio nuclide goes through the transformation chain. `""^118Cd underset(30min) rarr^(118)In underset(45min)rarr^(118)Sn ("stable")` The half-lives are written below the respective arrows. At time `t=0` only Cd was present. Find the fraction of nuclei transformed into stable over `60` minutes. |
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Answer» Correct Answer - A::C At time `t=t`, `N_1=N_0e^(-lambda_1t)` and `N_2=(N_0lambda_1)/(lambda_2-lambda_1)(e^(-lambda_1t)-e^(-lambda_2t))` `:.` `N_3=N_0-N_1-N_2` `=N_0[1-e^(-lambda_1t)-(lambda_1)/(lambda_2-lambda_1)(e^(-lambda_1t)-e^(-lambda_2t))]` `:.` `N_3/N_0=1-e^(-lambda_1t)-(lambda_1)/(lambda_2-lambda_1)(e^(-lambda_1t)-e^(-lambda_2t))` `lambda_1=0.693/30=0.0231 mi n^(-1)` `lambda_2=0.693/45=0.0154mi n^-1` and `t=60min` `:.` `N_3/N_0=1-e^(-0.0231xx60)-(0.0231)/(0.0154-0.0231)(e^(-0.0231xx60)-e^(-0.0154xx60))` `=1-0.25+3(0.25-0.4)` `=0.31` |
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| 302. |
Electrons accelerated from rest by a potential difference of 12.75V, are bombarded on a mono-atomic hydrogen gas. Possible emission of spectral lines are -A. first three Lyman lines, first two Balmer lines and first Paschen lineB. first three Lyman lines onlyC. First two Balmer lines onlyD. none of the above |
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Answer» Correct Answer - A |
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| 303. |
A `^7Li` target is bombarded with a proton beam current of `10^-4` A for 1 hour to produce `^Be` of activity `1.8xx10^8` disintegrations per second. Assuming that `^7Be` radioactive nucleus is produced by bombarding 1000 protons, determine its half-life. |
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Answer» Correct Answer - A::B::D At time t, let say there are N atoms of `^7Be` (radioactive). Then, net rate of formation of `^7Be` nuclei at this instant is `(dN)/(dt)=(10^-4)/(1.6xx10^-19xx1000)-lambdaN` or `(dN)/(dt)=6.25xx10^11-lambdaN` or `int_0^(N_0)(dN)/(6.25xx10^11-lambdaN)=int_0^3600dt` where, `N_0` are the number of nuclei at `t=1h` or `3600 s`. `:.` `-1/lambda1n((6.25xx10^11-lambdaN_0)/(6.25xx10^11))=3600` `lambdaN_0`=activity of `^7Be` at `t=1h=1.8xx10^8` disintegrations/s `:.` -1/lambda1n((6.25xx10^11-1.8xx10^8`)/(6.25xx10^11))=3600` `:.` `lambda=8.0xx10^-8sec^-1` Therefore, half-life `t_(1//2)=(0.693)/(8.0xx10^-8)=8.66xx10^6s` `=100.26` days |
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| 304. |
when Boron nuclus `(_(3)^(10) B)` is bombarded by neudrons , a- particle are emitted . The resulting nucleus is of the elenent …….. And has the mass different ……… are called isotopes . |
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Answer» Correct Answer - D `_(5)^(10) + _(0)^(1) rarr _(2)^(4) He + _(3)^(7) Li` The resulting nucleus is of element lithium and mass number is ` 7` . |
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| 305. |
In an ore containing uranium, the ratio of ^238U` to `206Pb` nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of `^238U`. Take the half-life of `^238U` to be `4.5xx10^9` years. |
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Answer» Correct Answer - A `(U^238)/(Pb^206)=3/1` `N_0=3+1=4` `N=3` `N=N_0e^(-lambdat)` …(i) `lambda=(1n2)/(t_(1//2)` …(ii) From Eqs. (i) and (ii), we get `t=1.88xx10^9 yr` |
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| 306. |
Find the increase in mass of water when `1.0 kg` of water absorbs `4.2 xx 10^3` J of energy to produce a temperature rise of `1K`. |
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Answer» Correct Answer - A::D `m=E/c^2=(4.2xx10^3)/((3.0 xx 10^8)^2)kg` `=4.7 xx 10^-14 kg` |
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| 307. |
A nucleus with mass number 220 initially at rest emits an `alpha`-particle. If the Q-value of the reaction is 5.5 MeV, calculate the kinetic energy of the `alpha`-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 Given that `K_1+K_2=5.5MeV` ...(i) From conservation of linear momentum, `p_1=p_2` or `sqrt(2K_1(216m))=sqrt(2K_2(4m))` as `p=sqrt(2km)` :. `K_2=54K_1`…(ii) Solving Eqs. (i) and (ii), we `K_2 = KE` of `alpha`-particle `=5.4MeV` :. The correct option is (b). |
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| 308. |
In the fusion reaction `_1^2H+_1^2Hrarr_2^3He+_0^1n`, the masses of deuteron, helium and neutron expressed in amu are 2.015, 3.017 and 1.009 respectively. If 1 kg of deuterium undergoes complete fusion, find the amount of total energy released. 1 amu `=931.5 MeV//c^2`. |
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Answer» Correct Answer - A::C `Deltam=2(2.015)-(3.017+1.009)=0.004`amu `:. Energy released `=(0.004xx931.5)MeV=3.726MeV` Energy released per deuteron `=(3.726)/(2)=1.863MeV` Number of deuterons in `1kg=(6.02xx10^26)/(2)=3.01xx10^26` :. Energy released per kg of deuterium fusion `=(3.01xx10^26xx1.863)=5.6xx10^26MeV` `=9.0xx10^13J` |
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| 309. |
`92^(U^(235)` mucleus absorbs a slow neutron and undergoes fission into `54^(X^139)` and `38^(sr^94)` nuclie The other particies produced in this fission process areA. `1beta` and `1alpha`B. `2 beta` and 1 neutronC. 2 neutronsD. 3 neutrons |
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Answer» Correct Answer - B |
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| 310. |
A sample of radioactive material has mass `m`, decay constant `lambda`, and molecular weight `M`. Avogadro constant `=N_(A)`. The initial activity of the sample is:A. `lamdam`B. `(lamdam)/(M)`C. `(lamdam N_(A))/(M)`D. `mN_(A)e^(lamda)` |
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Answer» Correct Answer - c |
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| 311. |
The activity of a radioactive substance is `R_(1)` at time `t_(1)` and `R_(2)` at time `t_(2)(gt t_(1))`. Its decay cosntant is `lambda`. Then .A. `R_(1)t_(1)=R_(2)t_(2)`B. `R_(1)=R_(1)e^(-(lambda^(t_(1)-t_(2))))`C. `(R_(1)-R_(2))/(t_(2)-t_(1))`=constantD. `R_(2)=R_(1)e^(lambda(t_(2)-t_(1))` |
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Answer» Correct Answer - B |
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| 312. |
What are the respective number of `alpha` and `beta`-particles emitted in the following radioactive decay?A. (a) 6 and 8B. (b) 6 and 6C. (c) 8 and 8D. (d) 8 and 6 |
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Answer» Correct Answer - D Let `n-alpha` particles and `m-beta` particles are emitted. Then, `90-2n+m=80` …(i) `200-4n=168` …(ii) Solving Eqs. (i) and (ii), we get `n=8` and `m=6` |
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| 313. |
`N_(1)` atoms of a radioactive element emit `N_(2)` beta partilces per second. The decay cosntant of the element is (in `s^(-1)`)A. `(N_(1))/(N_(2))`B. `(N_(2))/(N_(1))`C. `N_(1) ln (2)`D. `N_(2) ln (2)` |
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Answer» Correct Answer - B |
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| 314. |
If the half lives of a radioactive element for `alpha` and `beta` decay are 4 year and 12 years respectively, then the percentage of the element that remains after 12 year will beA. `6.25 %`B. `5.25 %`C. `4.25%`D. `3.50%` |
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Answer» Correct Answer - A |
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| 315. |
The mean lives of an unstable nucleus in two different decay processes are 1620 yr and 405 yr, respectively. Find out the time during which three-fourth of a sample will decay. |
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Answer» Correct Answer - D Let at some instant of time t, number of nuclei are N. Then, `((-dN)/(dt))_(net)=((-dN)/(dt))_1+((-dN)/(dt))_2` If the effective decay constant is `lambda`, then `lambdaN=Lambda_1N+lambda_2N` or `lambda=lamba_1+lambda_2=1/1620+1/405=1/324year^-1` Now, `N_0/4=N_0e^(-lambdat)` :. `-lambdat=1n(1/4)=-1.386` or `(1/324)t=1.386` :. `t=449 yr` |
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| 316. |
The graph represents the decay of a newly-prepared sample of radioactive nuclide X to a stable nuclide Y. The half-life of X is t. The growth curve for Y intersects the decay curve for X after time T. What is the time T ? A. `t//2`B. `ln (t//2)`C. `t`D. `ln (2t)` |
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Answer» Correct Answer - C |
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| 317. |
The half-lives of radioactive sample are 30 years and 60 years for two decay processes. If the sample decays by both the processes simultaneously. The time after which, only one-fourth of the sample will remain isA. (a) 10 yearsB. (b) 20 yearsC. (c) 40 yearsD. (d) 60 years |
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Answer» Correct Answer - C `lambda=lambda_1+lambda_2` `:.` `(1n2)/(T)=(1n2)/(T_1)=(1n2)/(T_1)` (T=Half-life) or `T=(T_1T_2)/(T_1+T_2)=20y` `1/4th` sample remains after 2 half-lives or 40 y. |
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| 318. |
A fraction `f_(1)` of a radioactive sample decays in one mean lie and a fraction `f_(2)` decays in one half-lifeA. `f_(1)gtf_(2)`B. `f_(1)ltf_(2)`C. `f_(1)=f_(2)`D. May be (a) ,(b) or (c ) depending on the values oof the mean life and half - life . |
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Answer» Correct Answer - a |
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| 319. |
A radioactive nuclide can decay simultaneously by two different processes which have decay constant `lambda_(1)` and `lambda_(2)`. The effective decay constant of the nucleide is `lambda`A. `lamda=lamda_(1)+lamda_(2)`B. `lamda=(1)/(2) (lamda_(1)+lamda_(2))`C. `(1)/(lamda)=(1)/(lamda_(1))+(1)/(lamda_(2))`D. `lamda=sqrt(lamda_(1)lamda_(2))` |
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Answer» Correct Answer - a |
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| 320. |
The wavelength of the spectral live the Balner series of hydrogen atom is `6561 A^(@)` . The wavelength of the second spectralline in the Balmer series of singly - ionized belium atom isA. `1215 A^(2) `B. `1640 A^(2) `C. `2430 A^(2) `D. `4687 A^(2) ` |
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Answer» Correct Answer - A we know that `(1)/(lambda) = RZ^(2) [(1)/(n_(1)^(2)) - (1)./(n_(2)^(2))] The wave length of spectal line in the balmar series of hydrogen atom is `6561Å` Here n_(2) = 3 and n_(1) = 2 ` ` :. (1)/(6561) = R(1)^(2) ((1)/(4) - (1)/(9)) = (5R)/(36) ` ....(i) For the second spectral line is the balmer of singly ionised belium ion `n_(2) = 4 and n_(1) = 2 , Z = 2 ` `:. (1)/(lambda) = R(2)^(2) [(1)/(4) - (1)/(16)] = (3R)/(4) `....(ii) Dividingh equation (i) and equqtion (ii() we get `(lambda)./(6561) = (5R)/(36) xx (4)/(3R) = (5)/(27)` ` :,. lambda= 1215 Å` |
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| 321. |
Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron . Apply the Bohr atom model and consider all possible transitions of this hypothetical particle that will be emitted level . The longest wavelength photon that will be emitted has longest wavelength `lambda` (given in terms of the Rydberg constant `R` for the hydrogen atom) equal to |
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Answer» In hydrogen atom `E_n = -(Rhc)/(N^2)` The longest wavelength `lambda_(max) ( or minimum energy) photon will correspond to the transition of particle form n= 3 to n =2. `:. `(hc)/(lambda_max) = E_3 - E_2 = 2Rhc ((1)/(2^2)) - (1)/(3^2))` This gives, `lambda_(max) = 18 //5R` `:. The correct option is (c ). |
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| 322. |
A nuclear of mass `M +deltam `is at rest and decay into two daughter nuclei of equal mass `(M)/(2)` each speed is `c` The binding energy per nucleon for the nucleus is `E_(1)` and that for the daugther nuclei is `E_(2)` ThenA. `E_(2) = 2E_(1)`B. `E_(1) gt E_(2)`C. `E_(2) gt E_(1)`D. `E_(1) = 2E_(2)` |
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Answer» Correct Answer - C In nuclear fesion , the binding energy per nucleon of doughter is graeter than the parent nuclear |
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| 323. |
The figure shows the variation of photocurrent with anode potential for a photosensitve surface for three different radiations. Let `l_a, l_b and l_c` be the curves a, b and c, respectively (a) `f_a = f_b and l_a != l_b` (b) `f_a = f_c and l_a = l_c` (c ) `f_a = f_b and l_a = l_b` (d) `f_b = f_c and l_b = l_c` |
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Answer» Correct Answer - A Saturation current is proportional to intensity while stopping potential increases with increases in frequency. Hence, `f_a = f_b while i_a lt l_b` Therefore, the correct option is (a). |
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| 324. |
Which of the following transition in `He^(+)` ion will give rise to a spectral line which has the same wavelength as some spectral line in the hydrogen atom ?A. n=4 to n=2B. n=6 to n=2C. n=6 to n=3D. n=8 to n=4 |
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Answer» Correct Answer - A::D |
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| 325. |
The ionization potential of H-atom is 13.6 V. The H-atoms in ground state are excited by mono chromatic radiations of photon energy 12.09 ev. Then the number of spectral lines emitted by the excited atoms, will beA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C |
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| 326. |
Canal rays are positively charged. |
| Answer» Correct Answer - 1 | |
| 327. |
An alpha particle of energy `5 MeV` is scattered through `180^(@)` by a found uramiam nucleus . The distance of closest approach is of the order ofA. `10^(-12) cm`B. `10^(-10) cm`C. `1A`D. `10^(-15) cm` |
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Answer» Correct Answer - A KEY CONCEPT : Distance of closest approach `r_(0) = (Ze(2e))/(4 pi epsilon _(0)E)` Energy `E = 5 xx 10^(6) xx 10^(-19) J` `:. R_(0) = (9 xx 10^(9) xx (92 xx 1.6 xx 10^(-19)) (2 xx 1.6 xx 10^(-19))/(5 xx 10^(6) xx 1.6 xx 10^(-19))` rArr r = 5.2 xx 10^(-14)m = 5.3 xx 10^(-12) cm` |
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| 328. |
Alpha particles are________charged. |
| Answer» Correct Answer - positively | |
| 329. |
X-rays travel at a speed of _________`ms^(-1)`. |
| Answer» Correct Answer - `3 xx 10^(8) ms^(-1)` | |
| 330. |
______rays are highly energized electrons. |
| Answer» Correct Answer - Cathode rays and beta | |
| 331. |
Beta rays emitted by a radicactive material areA. electrongnetic radiations.B. the electrons orbiting around the nucleus .C. charged particle emited by the nacleus .D. ineutral particles |
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Answer» Correct Answer - C (c ) `beta -` particle are changed particle emitted by the nucleus . |
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| 332. |
An X-ray tube is operated at 20 kV. The cut off wavelength isA. `0.89 Å`B. `0.75 Å`C. `0.62 Å`D. None of these |
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Answer» Correct Answer - C `lambda_(min) = (12375)/(V(in volts)) in Å` ` = (12375)/(20xx10^3 = 0.62 Å.` |
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| 333. |
`beta` rays are emitted from the _________. |
| Answer» Correct Answer - nucleus of an atom | |
| 334. |
`_86A^222rarr_84B^210`. In this reaction, how many `alpha` and `beta` particles are emitted?A. `6alpha , 3 beta`B. `3 alpha , 4 beta`C. `4 alpha , 3 beta`D. `3 alpha, 6 beta` |
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Answer» Correct Answer - B Ler `alpha` particles are n and `beta` particles are m. Then, `86-2n + m = 84 ….(i)` 222 - 4n =210 …(ii) Solving these two equations, we get n= 3 and `beta =4.` |
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| 335. |
The energy spectrum of `beta - particle` [number N€ as a function of `beta - energy E]` emitterfrom a radioactive source isA. B. C. D. |
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Answer» Correct Answer - C The range of energy of `beta` - particle is from zero to some maximum value |
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| 336. |
Statement I : Wavelength of continuous X-ray varies from a minimum value to infinity. Statement II : Continuous X-rays are emitted due to transition of electron from higher to lower energy level.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 - A |
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| 337. |
The curve representing the energy spectrum of `beta` -particles isA. B. C. D. |
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Answer» Correct Answer - A |
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| 338. |
Cut off potential for a metal in photoclectric effect for light of wavelength `lambda_(1), lambda_(2)` and `lambda_(3)` is found to be `V_(1),V_(2)` and `V_(3)` volts , If `V_(1),V_(2)` and `V_(3` are in Arithmetic progression then `lambda_(1),lambda_(2)` and `lambda_(3)` will be inA. Arithmetic progressionB. Geometric progressionC. Harmonic progressionD. None of these |
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Answer» Correct Answer - C |
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| 339. |
Which one curve is correct-A. B. C. D. None |
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Answer» Correct Answer - A |
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| 340. |
The wavelengths and frequencies of photons in transition 1,2 and 3 for hydrogen like atom are `lambda_(1),lambda_(2),lambda_(3), v_(1),v_(2)` and `v_(3)` respectively. Then: A. `v_(3)=v_(1)+v_(2)`B. `v_(3)=(v_(1)v_(2))/(v_(1)+v_(2)`C. `lambda_(3)=lambda_(1)+lambda_(2)`D. `lambda_(3)=(lambda_(1)lambda_(2))/(lambda_(1)+lambda_(2)` |
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Answer» Correct Answer - A::D |
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| 341. |
In an X-ray tube if the electrons are accelerated through `140KV` then anode current obtained is 30mA. If the whole energy of electrons is converted into heat then the rate of production of heat at anode will beA. 968 calorieB. 892 calorieC. 1000 calorieD. 286 calorie |
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Answer» Correct Answer - C |
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| 342. |
If `lambda_(K_(alpha)), lambda_(K_(beta))` and `lambda_(L_(alpha))` are the wavelengths of `K_(alpha), K_(beta)` and `L_(alpha)`, lines respectively , thenA. `lambda_(K_(beta)) = (lambda_(K_(alpha)) lambda_(L_(alpha)))/(lambda_(K_(alpha)) + lambda_(L_(alpha)))`B. `lambda_(L_(alpha)) = (lambda_(K_(alpha)) lambda_(K_(beta)))/(lambda_(K_(alpha)) + lambda_(K_(beta)))`C. `lambda_(L_(alpha)) = (lambda_(K_(alpha))lambda_(K_(beta)))/(lambda_(K_(beta)) + lambda_(K_(alpha)))`D. none of these |
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Answer» Correct Answer - A |
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| 343. |
If the frequency of `K_(alpha), K_(beta)` and `L_(alpha)` , X-ray lines of a substance are `v_(K_(alpha)), v_(K_(beta))`, and `v_(L_(beta))`A. `v_(K_(alpha)) + v_(K_(beta)) = v_(L_(alpha))`B. `v_(K_(alpha)) - v_(K_(beta)) = v_(L_(alpha))`C. `v_(K_(alpha)) + v_(L_(alpha)) = v_(K_(beta))`D. none of these |
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Answer» Correct Answer - C |
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| 344. |
photons of energy `4.25 eV` strike the surface of metal A, the ejection photoelectric have maximum kinetic energy `T_(A) eV energy 4.70 eV is T_(B) = (T_(A) - 1.50) eV` if the de Brogle wavelength of these photoelec tron is `lambda_(B) = 2 lambda_(A) `, thenA. The work function of `A is 2.25 eV`B. The work function of `B is 4.20 eV`C. `T_(A) = 2.00eV`D. `T_(B) = 2.75eV` |
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Answer» Correct Answer - A::B::C `4.25 = W_(A) +T_(A)`….(i) `also T_(A) = (1)/(2) m nu_(A)^(2) = (1)/(2) (m^(2) nu_(A)^(2))/(m) = (p_(A)^(2))/(2m) = (h^(2))/(2m lambda_(A)^(2))` …..(ii) `[:.lambda = (h)/(p)]` for metal B `4.7 = (T_(A) - 1.5) + W_(B)` ….(iii) `also T_(B) = (h^(2))/(2mlambda_(B)^(2)) `....(iv) [as eq(ii)] Dividing equation (iv) by(ii) `(T_(B))/(T_(A)) = h^(2))/(2mlambda_(B)^(2)) xx (2m lambda_(A)^(2))/(h^(2)) = (lambda_(A)^(2))/(ambda_(B)^(2))` `rArr (T_(A) - 1.5)/(T_(A)) = (lambda_(A)^(2))/((2 lambda_(A))^(2)) = lambda_(A)^(2))/(4lambda_(A)^(2)) = (1)/(4)` `[ :. lambda_(B) = 2 lambda_(A )given ]` `rArr 4T_(A) - 6 = T_(A) rArr T_(A) = 2 eV` `from (i) `W_(A) = 2.25 eV ` `from (ii) W_(B) = 4.2 eV ` also` T_(B) = T_(A) - 1.5 rArr T_(B) = 0.5 eV` |
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| 345. |
which of the following statement (s) is (are) correct ?A. The rest mass of a stable nucleus is less then the sum of the rest mases opf its separated niuceonsB. The rest mass of a stable nucleus is greater then the sum of the rest mases opf its separated niuceonsC. in nuclear fission , energy is released by fasing two nuclei of medium mass (approximately `100amu`)D. in nuclear fission , energy is released by fregmentation of a very heavy nucleus |
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Answer» Correct Answer - A::D are correct option |
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| 346. |
Holes are charge carriers inA. intrinsic semicondutorsB. ionic solidsC. p- type samiondactorsD. inctals |
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Answer» Correct Answer - B::C Holes are electron vacancies which participate in electron conductivity These are producted in semiconductors |
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| 347. |
The half-life of `^215At` is `100mus`. The time taken for the activity of a sample of `^215At` to decay to `1/16th` of its initial value isA. (a) `400 mus`B. (b) `63mus`C. (c) `40 mus`D. (d) `300 mus` |
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Answer» Correct Answer - A `R=R_0(1/2)^n` …(i) Here, R= activity of radioactive substance after n half-lives `=R_0/16` (given) Substituting in Eq. (i), we get `n=4` `:.` `t=(n)t_(1//2)=(4)(100mus)=400mus` |
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| 348. |
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. (a) 3.8 daysB. (b) 16.5 daysC. (c) 33 daysD. (d) 76 days |
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Answer» Correct Answer - B Using `N=N_0e^(-lambdat)` where, `lambda=(1n2)/(t_(1//2))=(1n2)/(3.8)` `:.` `N_0/20=N_0e^(-(1n2)/(3.8)t)` Solving this equation with the help of given data we find `t=16.5 days` `:.` Correct option is (b). |
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| 349. |
Let `m_(p)` be the mass of a poton , `M_(1)` the mass of a `_(10)^(20) Ne` nucleus and `M_(2)` the mass of a `_(20)^(40) Ca` nucleus . ThenA. `M_(2) = 2M_(1)`B. `M_(2) gt 2M_(1)`C. `M_(2) lt 2M_(1)`D. `M_(1) lt 10(m_(n) + m_(p))` |
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Answer» Correct Answer - C::D KEY CONCEPT : Due to mass defact (which is finally reponsible of the binding energy of the mucleus ) mass of a mecleus is always less then the sum of masses of its constiluent particles `_(10)^(20)` Ne is made up of `10` protons `10` neutron therefore , mass of`_(10)^(20)` Ne nucleus `m_(1) lt 10(m_(p) + m_(n))` |
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| 350. |
The half - liofe of` ^(131) 1 `is days Given a sample of`^(131) 1` at time `t = 0` , we get thatA. no nucleus will decay `t = 4 days`B. no nucleus will decay `t = 8 days`C. no nucleus will decay `t = 16 days`D. a given nucleus may dacay at any time after `t = 0` |
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Answer» Correct Answer - D The result follow from the formula based on have of radioactive decay `N = N_(0)e^(- lambda1)` The nucleus start after time `1 = 0` |
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