InterviewSolution
Saved Bookmarks
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.
| 1251. |
Assertion `(A) : beta-` particles are deflected more than `alpha-` particles in a given electric field. Reason `(R) : `Charge on `alpha-` particles is larger than on `beta-`particles.a)If both `(A)` and `(R)` are correct , and `(R)` is the correct explanation of `(A)`b)If both `(A)` and `(R)` are correct, but (R) is not the correct explanation of `(A)`c)If `(A)` is correct, but `(R)` is incorrect.d)If `(A)` is incorrect, but `(R)` is correct. |
|
Answer» Correct Answer - Ture |
|
| 1252. |
The ration of spectific charge ( e//m) of a proton to that of an alpha-particle is :A. `1 :4`B. ` 1:2`C. ` 1: 1//4`D. ` 2:1` |
|
Answer» Correct Answer - D `e//m` "for proton" ` = (1)/(1)` , `e//m` "for alpha-"particle" ` = (2)/(4)` . |
|
| 1253. |
`1.0 g` of `Mg "atom" ("atomic mass" = 24.0 "amu")` in the vapour phase absorbs `50.0 kJ` energy .Find the composition of the final maximum if the first and the second if of Mg are `740 kJ "mol^(-1)` respectively |
|
Answer» Mole of Mg `= 1//24` `= 4.167 xx 10^(-2) mol` Energy used in ionising Mg to `Mg^(o+)` `= (4.167 xx 10^(-2)) xx (740 kJ mol ^(-1))` `= 30.84 kJ` Energy used in ionising `Mg^(o+)` to `Mg^(2+)= (50.0 - 30.84 )kJ =- 19.16 kJ` Amount of `Mg^(o+)`converted to `Mg^(2+)` `= (19.16 xx kJ)/(1450 kJ mol^(-1)) 1.321 xx 10^(-2) mol` Amount of `Mg^(o+)` remaining as each `(4.167 xx 10^(-2) - 1.327 xx 10^(-2)) mol = 2.846 xx 10^(-2) mol` % of `Mg^(o+) = (2.846 xx 10^(-2))/(4.167 xx 10^(-2)) xx 100 = 68.3` % of `Mg^(2+) = 100 - 68.3 = 31.7` |
|
| 1254. |
Which of the following is associated with a de Brogle wave of longer wavelength -a proton or an electron having same velocity ? |
|
Answer» Correct Answer - A Wavelength of the de Broglie wave associated with a particle is inversely proporinal to the mass of the particle. Since the mass of electron is less than that of proton , therefore wavelength of the de Broglie waves associated with electron would be more |
|
| 1255. |
In a given electric field , the `beta` particle are deflected more than the `alpha` -particle in spin of the `alpha`- particle having a langer charge |
| Answer» In agiven electric field , the `beta` particle are deflected large than the `alpha` particle in spinof the `alpha` particle having large charge due to the higher value of e//m | |
| 1256. |
Which of the following is associated with a de Brogle wave of langer wavelength -a proton or an electron having same velocity ? |
| Answer» The wavelength of the de Brogle wave associated with a particle is inversel prepotionaly to the mass than that of proton wavelength of the de Broglie waves associated with the electron will be more | |
| 1257. |
What are isobars? Are the physical and chemical properties of isobars similar ? |
|
Answer» The atoms of different elements with different atomic numbers but having the same atomic mass are called isobars . The isobars have different physical and chemical properties . |
|
| 1258. |
An atom of an element has two electrons in its outermost shell. Find the nature and value of charge on the ion produced when these tow electrons are lost . |
| Answer» The ion will carry double positive charge | |
| 1259. |
An atom has 2 electrons in K shell, 8 electrons in L shell and 6 electrons in M shell. The number of s-electrons present in that element isA. 6B. 5C. 7D. 10 |
|
Answer» Correct Answer - A Shell =K,L,M =`1s^2 2s^2 2p^6 3s^2 3p^4` Hence the number of s electron is 6 in that element |
|
| 1260. |
Which of the following statement is false ?A. Violet radiations have a longer wavelength than red radiationsB. The energy of light with `lambda=600nm` is lower than that of `lambda=500nm`C. Spectrum of an atom is known as line spectrumD. The wavelength associated with an electron is longer than that of proton if they have the same speed |
|
Answer» Correct Answer - A Voilet radiations have a lower wavelength than red radiations. |
|
| 1261. |
If the speed of electron in the first bohr orbit of hydrogen atom is x then the speed of the electron in the third Bohr orbit of hydrogen isA. `x//9`B. `x//3`C. `3x`D. `9x` |
|
Answer» Correct Answer - B `V_(n) = (V_(1))/(n)` |
|
| 1262. |
Energy of electron of hydrogen atom in second Bohr orbit isA. `-5.44 xx 10^(-19) J`B. `-5.44 xx 10^(-19) kJ`C. `-5.44 xx 10^(-19) ` calD. `-5.44 xx 10^(-19) ` eV |
|
Answer» Correct Answer - A `E=-(2.172xx10^(-18))/n^2=-(2.172xx10^(-18))/2^2` `=-5.44xx10^(-19)` J |
|
| 1263. |
How much energy is required to ionise a H atom if the electron occupies `n = 5` orbit?A. `5.44ev`B. `10.8ev`C. `0.544ev`D. `1.08ev` |
|
Answer» Correct Answer - C `-E_(n) = I.E. = (13.6)/(n^(2)) xx z^(2)` |
|
| 1264. |
How much energy is required to ionise a H - atom if the electron occupie ` n=5 ` orbit ? Compare your answer with the ionization energy fo H-atom (energy required to remove the electron from ` n= 1 ` orbit ). |
|
Answer» ` E_1 =- 13. 6 eV,` ` E_5 = ( - 13. 6)/(5^2) =- 0. 544 eV` ` =- 0.544 xx 1.6022 xx 10^(-19) J =- 8. 71 xx 10^(-20) J` Thus , ` 0.544 eV` energy is required to ionise H-atom of electron is in ` 5^(th)` orbit . Also , ` E_5 /E_1 = (0.544 )/( 13.6) = 0.04`. |
|
| 1265. |
How is the radius of a nucleus related to its mass number ?A. `R=R_O A^(1//2)`B. `R=R_O A`C. `R=R_O A^2`D. `R=R_O A^(1//3)` |
|
Answer» Correct Answer - D `R=R_0(=1.4xx10^(-13) cm )xxA^(1//3)` |
|
| 1266. |
Probability of finding an electron at the nodal surface isA. UnityB. LowC. HighD. Zero |
|
Answer» Correct Answer - D Electron density is zero at nodal region |
|
| 1267. |
Which of the following is not a fundamental particleA. ProtonB. NeutronC. `alpha`-particleD. Electron |
|
Answer» Correct Answer - C electrons, protons and neutrons are fundamental particulars |
|
| 1268. |
Energy required to ionise 2 mole of gaseous `He^(+)` ion present in its ground state is :A. `54.4 eV`B. `108.8 N_(A)eV`C. `54.4 N_(A)eV`D. `108.8eV` |
|
Answer» Correct Answer - B `I.E = +13.6 xx (Z^(2))/(n^(2))eV = 13.6 xx 4 = 54.4 eV` for male `= 54.5 xx 2 xx N_(a)eV = 108.8 N_(A)eV` |
|
| 1269. |
The radius of the first orbit of H-atom is r. then the radius of the first orbit of `Li^(2+)` will be:A. r/9B. r/3C. 3rD. 9r |
|
Answer» Correct Answer - B Formula for radius of H-like species, `r_n=0.529xxn^2/Z Å` As `0.529xxn^2` is radius for H atom which is equal to r (given ) So, `r_n=r/3` (n=1, Z=3, for `Li^(2+)` ) |
|
| 1270. |
Correct order of radius of the first orbit of `H, He^+, Li^(2 +), Be^(3 +)` is :A. `H gt He^(+) gt Li^(2+) gt Be^(3+)`B. `Be^(3+) gt Li^(2+) gt He^(+) gt H`C. `He^(+) gt Be^(3+) gt Li^(2+) gt H`D. `He^(+) gt H gt Li^(2+) gt Be^(3+)` |
| Answer» Correct Answer - A | |
| 1271. |
The de-Brogile wavelength for a proton with a velocity `15%` of the speed of light is:A. `8.8 xx 10^(-12)m`B. `8.8 xx 10^(-14)cm`C. `8.8xx 10^(-15)m`D. `4.4 xx 10^(-15)cm` |
|
Answer» Correct Answer - C `lambda = (h)/(mv)` |
|
| 1272. |
If the following mater travel with equal velocity the longest wavelength is that ofA. ElectronB. `alpha`-particleC. ProtonD. Neutron |
|
Answer» Correct Answer - A `lambda prop (1)/(m)` |
|
| 1273. |
If the following mater travel with equal velocity the longest wavelength is that ofA. electronB. protonC. neutronD. particle |
|
Answer» Correct Answer - A `lambda = (h)/(mv)` |
|
| 1274. |
Wavelength of the first line of Paschen series is `- (R = 109700 cm^-1`)A. `[18750 Å]`B. `[2854 Å]`C. `[3452 Å]`D. `[6243 Å]` |
|
Answer» Correct Answer - A (a) `[18750 Å]`. |
|
| 1275. |
For the Paschen series thr values of `n_(1)` and `n_(2)` in the expression `Delta E = R_(H)c [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` areA. `n_(1)=1,n_(2)=2,3,4` . . .B. `n_(1)=2,n_(2)=3,4,5` . . .C. `n_(1)=3,n_(2)=4,5,6` . . .D. `n_(1)=4,n_(2)=5,6,7` . . |
| Answer» Correct Answer - C | |
| 1276. |
Which of the following has the largest de Brogile wavelength provided all have equal velocity?A. Carbon dioxide moleculeB. Ammonia moleculeC. Oxygen moleculeD. Nitrogen molecule |
|
Answer» Correct Answer - B `lambda = (h)/(mv), lambda prop (1)/(m)` |
|
| 1277. |
If the following mater travel with equal velocity the longest wavelength is that of a //amA. ElectronB. ProtonC. NeutronD. `alpha` particle |
|
Answer» Correct Answer - A `lambda = (h)/(mv)` `lambda` will be if m is small |
|
| 1278. |
First line of Paschen series has wave number `(R_(H)=109700//cm)`A. `2854xx10^(8)(Å)^(-1)`B. `6243xx10^(8)(Å)^(-1)`C. `6856xx10^(8)(Å)^(-1)`D. `3452xx10^(8)(Å)^(-1)` |
|
Answer» Correct Answer - C `n_(l)=3 n_(f)=4` `1/lambda=109677[1/n_(i)^(2)-1/n_(f)^(2)]=6856xx10^(8)(Å)^(-1)` |
|
| 1279. |
The highest excited state that unexcited hydrogen atom can ereach when they are bombarged with ` 12. 2 eV` electron is :A. ` n=1`B. ` n=2`C. ` n=3`D. ` n=4` |
|
Answer» Correct Answer - C `E_(1) = - 13.6eV` After absorpton fo ` 12. 2 eV` energy ` E_(H) = - 13 . 6 + 12 . 2 =- 14 eV` Now ` E_(n) = E_(t) /n^(2) :. N^(2) = ( - 13 . 6 )/( -1.4) = 9.71` `therefore n=3` . |
|
| 1280. |
When greater number of excited hydrogen atoms reach the ground state, thenA. More number of lines are found in Lyman seriesB. The intensity of lines in Balmer series increaseC. The intensity of lines in Lyman series increaseD. Both the intensity and number of lines in Lyman series increases |
|
Answer» Correct Answer - C intensity `alpha` no of photons |
|
| 1281. |
The highest excited state that an unexcited hydrogen atom can reach when they are bombarded with ` 12.2 eV` electron is : |
|
Answer» Correct Answer - 3 `E_1=-13.6` eV After absorption of 12.2 eV energy =-1.4 eV Now `E_n=E_1/n^2 therefore n^2=(-13.6)/(-1.4) =9.71` `therefore n=3` |
|
| 1282. |
In aufbau principle, the word aufbau represents :A. The name of scientistB. German term meaning for buliding upC. The enrgy of elerctronD. The angular momentum of electron |
|
Answer» Correct Answer - B It is a fact . |
|
| 1283. |
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition `n = 4` to `n=2` of ` He^(o+)` spectrum? |
|
Answer» For ` He^+ : 1/ (lambda) = R _H xx Z^2 [1/n_1^2 - 1/n_2^2]`, ` 1/ (lambda)R_H xx 4 [ 1/2^2 - 1/4^2 ] ( :. Fo r He Z=2) …(1)` For H -atom `: (1)/ (lambda)R_H [1/n_1^2 - 1/n_2^2]` …(2) For same ` (lambda)`,by eqs. (1) and (2), ` therefore 2^2 [ 1/2^2 - 1/4^2] = [ 1/n_1^2 - 1/n_2^2] , [ 1/1^2 - 1/2^2] = [ 1/n_1^2 - 1/n_2^2] ` `therefore n_1 = 1 "and " n_2 =2`. |
|
| 1284. |
When greater number of excited hydrogen atoms reach the ground state :A. More number of lines are found in Lyman seriesB. The ines in Balmer seies increasesC. The intensity of lines in Lyman seies increaseD. Both the intensity and number of lines in Lyman series increased . |
|
Answer» Correct Answer - C Intensity , i.e., no. of photons emitted out per sec . Increases . |
|
| 1285. |
The probability of finding elercrtron at a distance r form the nucleus for H or H like atom is :A. `psi`B. `1/(pi) [Z/a_0 ]^3 .e^(-2zr)/(a_0)`C. `1/(pi) [Z/a_0 ]^(3//2) -e^(zr)/a_0`D. None of these |
|
Answer» Correct Answer - B Probabuility for finding electron `= psi^2` and `psi = 1/(sqrt pi) [ Z/a-0 ]^(3//2) .d^(-Z2//a-)` |
|
| 1286. |
Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength `975A^(@)`. How many different lines are possible in the resulting spectrum? Calculate the longest wavelength amongst them. |
|
Answer» Correct Answer - six, `18800A^(@)` `lambda=975Å` `E=(lambdac)/(lambda)=(6.626xx10^(-34)xx3xx10^(8))/(975xx10^(10))=2.03xx10^(-18)J=12.75eV` So electron will excite to 4th energy level and when comeback number of emission line will be 6. minimum energy emitted `=E_(4)-E_(3)=0.66eV` `lambda=(hc)/(E)=(1.9878xx10^(-25))/(66xx1.6xx10^(-19))=1.882xx10^(-6)m=18820Å` |
|
| 1287. |
What is the number fo photons fo light with a wavelngth fo ` 400 p m` that provide ` 1 J ` of energy ? |
|
Answer» ` E//"photon" = (hc)/ ( lambda) = (6.625 xx 10^(-34) xx 3 . 0 xx 10^8)/( 4000 xx 10^(-12))` ` = 4 . 97 xx 10^(-17)J` `therefore` No . of photon in 1 J energy `= 1/( 4.97 xx 10 ^(-17))` ` 2. 01 xx 10 ^(16) ` photons . |
|
| 1288. |
The number of waver made by an elercrtron moving in an orbit having maximum magetic quantum number `+ 3` is :A. `4`B. `3`C. `5`D. `6` |
|
Answer» Correct Answer - A If `m=3` (maxumum), then `l =3` and n can have maximum value of four . Also no. of waves in n orbit = no or orbit . |
|
| 1289. |
Match the column: |
| Answer» Correct Answer - (A)P(B)P,Q,R,S,T(C)Q,R,S(D)P | |
| 1290. |
The ............ on a proton is equal in magnitude but opposite in sign to that of an ............. |
| Answer» Charge , electrons | |
| 1291. |
Isotopes have the same .............. .......... but different .......... ............ |
| Answer» atomic number , atomic masses | |
| 1292. |
orbital isA. Circular path around the nucleus in which the electron revolvesB. Space around the nucleus where the probability of finding the electron is maximumC. Amplitude of electrons waveD. None of these |
| Answer» Correct Answer - B | |
| 1293. |
For an electron if the uncertainty in velocity is `Deltav`, the uncertainty in its position `(Deltax)` is given byA. `(hm)/(4piDeltav)`B. `(4pi)/(hmDeltav)`C. `h/(4pimDeltav)`D. `(4pi m)/(h.Deltav)` |
| Answer» Correct Answer - C | |
| 1294. |
The wave nature of an electron was first given byA. de-BroglieB. HeisenbergC. MosleyD. Sommerfield |
| Answer» Correct Answer - A | |
| 1295. |
The wave charaters of electron was experimenally verified by ………..A. BohrB. De-BroglieC. Davison and GermerD. Schrodinger |
| Answer» Correct Answer - C | |
| 1296. |
What should be the mass of the sodium photon if its wavelength is `5894Å`, the velocity of light is `3xx10^(8)` metre//second and the value of `h` is `6.6252xx10^(-34)kg m^(2)//sec`. ?A. `3.746xx10^(-26)`B. `3.746xx10^(-30)`C. `3.746xx10^(-34)`D. `3.746xx10^(-36)` |
|
Answer» `lambda=(h)/(mxxc) implies m=(h)/(clambda)` `(because lambda=5894Å=5894xx10^(-10)m)` `m=(6.6252xx10^(-34))/(3xx10^(-8)xx5894xx10^(-10)) " "or" "(6.652)/(17682)xx10^(-32)` `=0.0003746xx10^(-32)=3.746xx10^(-36)kg` |
|
| 1297. |
The wave character of moving electron was experimentally verified by :A. de-BroglieB. N BohrC. Davisson and GermerD. Schrodinger |
|
Answer» Correct Answer - C (c) The wave nature of an electron is proved by Davisson and Germer experiment. In this experiment the scattering pattern of an electron is similar to that of X-rays. |
|
| 1298. |
Hydrogen nucleus combines to form helium then energy is released. Binding energy/nucleon of `He` is greater than hydrogen.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If assertion is false but reason is true. |
|
Answer» Correct Answer - A (a) A nuclear reaction in which two lighter nuclei are fused together to form a heavier nuclei is called nuclear fusion. In such a process, more stable nuclei come into existence and binding energy per nucleon increases. Fusion reactions are highly exothermic because of less of mass during fusion. e.g. `._1^2 H + ._1^(2) H rarr ._2^(4) He + 24.9 MeV`. |
|
| 1299. |
The wave charaters of electron was experimenally verified by ……….. |
| Answer» Davisson and germer | |
| 1300. |
Consider the ground state `Cr` atom `(Z = 24)`. The number of electron with the azimuthal number `l = 1` and `2` ,respectively areA. 16 and 4B. 12 and 5C. 12 and 4D. 16 and 5 |
|
Answer» Correct Answer - B `._24Cr to 1s^2 , 2s^2 , underset"l=1"(2p^6) ,3s^2, underset"l=1"(3p^6), underset(l=2)(3d^5), 4s^1` (We know that for p the value of l=1 and for d, l=2) For l=1 total number of electron =12 For l=2 total number of electron =5 |
|