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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.
| 351. |
For an element, Z = 9. The valency of this element will be :A. 4B. 2C. 1D. 3 |
| Answer» Correct Answer - C | |
| 352. |
If Z = 3, what would be the valency of the element? Also, name the element |
| Answer» The symbol Z stands for the atomic number of an element. By saying that Z = 3, we mean that the atomic number of this element is 3. The electronic configuration of the element having atomic number 3 is `underset("2, 1")("K L")`. It has 1 electron in its outermost shell (L shell). So, one atom of this element can lose 1 electron to achieve the nearest inert gas electron arrangement of `underset(2)(K)` (which is the same as that of helium gas). Since one atom of this element loses 1 electron to achieve the inert gas electron configuration, thereforem its valency is 1. The element having atomic number 3 is lithium. | |
| 353. |
What valency will be shown by an element having atomic number 15 ? |
| Answer» The atomic number of this element is 15, so its electronic configuration will be `underset("2, 8, 5")("K L M")`. It has 5 electrons in its outermost shell, so it needs 3 more electrons to complete the eight electron, inert gas configuration. Since one atom of this element needs 3 electrons to achieve the inert gas electron configuration, therefore, its valency is 3. (The element having atomic number 15 is actually ? phosphorus). | |
| 354. |
The isotopes of an element contain :A. same number of neutrons but different number of protonsB. same number of neutrons but different number of electronsC. different number of protons as well as different number of neutronsD. different number of neutrons but same number of protons |
| Answer» Correct Answer - D | |
| 355. |
The ratio of mass of an electron to that of the mass of hydrogen atom isA. `1 : 3871`B. `1: 1837`C. `1: 1296`D. ` 1: 3781` |
| Answer» Correct Answer - 2 | |
| 356. |
Which of the following is a correct statement ?A. Anode rays arise from anode ( so called)B. Anode rays are also known as canal raysC. The nature of anode rays depend upon the nature of gas in the discharge tubeD. All of these |
| Answer» Correct Answer - 4 | |
| 357. |
Which of the electromagnetic waves result from charges within the nucleus of the atom?A. Long radio wavesB. Visible light wavesC. `X` raysD. Gamma rays |
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Answer» Correct Answer - D The long radiowaves are emitted by large antennas such as those used by broadcasting stations. Visible light waves are produced by the motion of electrons within atoms and molecules. The most common source of `X` rays is the deceleration of high-enegry electrons bombarding a metal traget. |
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| 358. |
When alpha particle are sent through a thin metal foil ,most of them go straight through the foil becauseA. alpha particles are much heavier than electronsB. alpha particle are positively chargedC. most part of the atom is empty spaceD. alpha particles move with high velocity |
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Answer» Correct Answer - A::C are correct options. |
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| 359. |
How many times is an atom bigger than the nucleus ?A. 100000B. 5000C. 10000D. 200 |
| Answer» Correct Answer - 1 | |
| 360. |
What `alpha`-rays strikea thin gold foil thenA. Most of the `alpha-`rays do not pass through the gold foilB. Most of the `alpha-` rays get deflected backC. Most of the`alpha-`rays get deflected through small anglesD. Most of the `alpha-`rays pass through without any deviation. |
| Answer» Correct Answer - 4 | |
| 361. |
Canal rays areA. a strem of positronsB. a stream of protonsC. a stream of positively charged ionsD. electromagentic waves |
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Answer» Correct Answer - C Anode rays are called canal rays because they pass through the canals, i.e., holes in the cathode. Anode rays are made up to positively changed ions because they are produced as a result of the knockout of electrons form the atoms of the gas by the bombardment of high-speed electrons of the cathode rays on them. |
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| 362. |
Which of the following metaks has the highest value of exchnage enegry?A. ZnB. `Cr`C. `Cu`D. `Mn` |
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Answer» Correct Answer - C The electronic configuration of `Cu` is `[Ar]3d^(10)4d^(1)`. It has completely filled `3d` subshell which can hev exchanges of electrons pairs as well as of individual electrons resulting in maximum number of exchanges and, hence, maximum exchange energy. |
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| 363. |
Which of the following transitions will have minimum wavelength ?A. `n_(4) rarr n_(1)`B. `n_(2) rarr n_(1)`C. `n_(4) rarr n_(2)`D. `n_(3) rarr n_(1)` |
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Answer» Correct Answer - A `Delta E = hv = h (c)/(lamda) or lamda (hc)/(Delta E), " i.e., " lamda prop (1)/(Delta E). Delta E = E_(4) - E_(1)` is maximum |
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| 364. |
Assertion. According to Thomson model of atom, mass of the atom is concentrated in the centre of the atom. Reason. According to Thomson model, positive charge is concentrated in the centre of the atomA. If both assertion and reason are true, and reason is the true explanation of the arrertionB. If both assertion and reason are true, but reason is not the true explanation of the assertionalC. If assertion is true, but reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - D Correct A. According to Thomson model of atom, mass of the atom is uniformly distributed over the atom. Correct R. According to Thomosn model, positive charge is uniformly distributed over the atom. |
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| 365. |
Which of the following is the energy of a possible excited state of hydrogen?A. `-3.4 eV`B. `+ 6.8 eV`C. `+13.6 eV`D. `-6.8 eV` |
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Answer» Correct Answer - A The energy of electron in the excited state `(n gt 1)` is negative. By taking energy as-3.4 `eV_(i)`value of n is calculated as : `n=sqrt((-13.6)/("Eexcited state"))=sqrt((-13.6)/(-3.4))=sqrt(4)=2` `:.` It is the correct answer. |
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| 366. |
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Find the symbol of the element. |
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Answer» Mass number of the element (p+n)=81 Let the number of protons =`x` `:.` Number of neutrons present `=x+(31.7)/(100) xx x =1.317x` `x+1.317 x=81 or 2.317x=81` `x=(81)/(2.317)=35` The element with 35 electrons (or Z=35) is bromine (Br) |
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| 367. |
Is an electron stationary in stationary energy state ? |
| Answer» No. the electron is not stationary. Only the energy asspciated with it is stationary. | |
| 368. |
Can a moving cricket ball have a wave character ? |
| Answer» According to de Broglie relation, `lambda=h//mv`. The mass of the cricket ball (macroparticle) is sufficiently large. Therefore , its wavelength will be very short and it will not be possible to measure the same. Thus, a moving cricket ball cannot have wave character. | |
| 369. |
What is the significance of de Broglie relationship ? |
| Answer» The de Broglie relationship regarding the dual nature of matter is applicable only to the moving microscopic particles. For semi-micro and the macro particles, the particle character is very large and the wave character is very small. In other words, the wavelength associated with such particles is so small that it cannot be measured by any of the available methods. | |
| 370. |
Two particles A and B are in motion. If the wavelength associated with particle A is `5 xx 10^(-8)m`, calculate the wavelength associated with particle B if its monentum is half of A |
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Answer» By de Broglie equation, `lamda_(A) = (h)/(p_(A)) and lamda = (h)/(p_(B))` `:. (lamda_(A))/(lamda_(B)) = (p_(B))/(p_(A))` But `p_(B) = (1)/(2) p_(A)` (Given) `:. (lamda_(A))/(lamda_(B)) = (1//2p_(A))/(p_(A)) = (1)/(2) or lamda_(B) = 2xx lamda_(A) = 2 xx 5 xx 10^(-8) m = 10^(-7)m` |
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| 371. |
Which one `Fe^(3+), Fe^(2+)` is more paramagnetic and why ? |
| Answer» As `Fe^(3+)` contains 5 unpaired electrons while `Fe^(2+)` contains only 4 unpaired electrons, `Fe^(3+)` is more paramagnetic. | |
| 372. |
A hypothetical electromagnetic wave is shown in figure. Find out the wavelength of the radiation. |
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Answer» Wavelength in the distance between two successive peaks or two successive throughs of a wave. Therefore, `lambda=4xx2.16pm=8.64pm` `=8.64xx10^(-12) [because 1"pm"=10^(-12)m]` |
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| 373. |
In photoelectric effect light of certain frequency `( gt ` threshold frequency ) is incident on a metal surfave whereby , an `e^(-)` ( with certain K.E.) moves towards the collector plate and a flow of current is initiated . In order to stop the current flow, an opposite potential , on the two metal plates, is applied The work function of metal is 6eV . If light of frequency ` 1 xx 10^(15) `Hz is incident on the metal , intensity of light is increased 4 times, thenA. No photoelectron will be ejectedB. 8 photoelectrons of zero kinetic energy shall be ejectedC. 2 photoelectrons of 2 eV kinetic energy are ejectedD. Only one photoelectron is ejected |
| Answer» Correct Answer - 1 | |
| 374. |
What type of metals are used in photoelectric cells ? Give one example. |
| Answer» The metals with low ionisation enthalpies are used in photoelectric cells. Cesium (Cs) on alkali metal belonging to group 1 is the most commonly used metal. | |
| 375. |
What will happen to the mass of electron if it travels with velocity of light ? |
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Answer» When the electron is moving with velocity `upsilon`, its mass is given as `(m) =(m_(0))/((1-(upsilon^(2))/(c^(2)))^(1//2)` Here `" " m_(0)`=rest mass of the electron . If If `upsilon`=c , then `m=(m_(0))/((1-(c^(2))/(c^(2)))^(1//2)=(m_(0))/((1-1))1//2)=(m_(0))/(0) or m = oo` Thus, the mass of electron moving with velocity of light if infinity. |
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| 376. |
In summer , we are advised not to wear black clothes. Assign reason. |
| Answer» We know that the black bodies are good absorbers and they absorb all the radiations which are falling on them. Thus, if one wears black clothes, he or she is likely to feel more hot in summer. | |
| 377. |
According to de-Brogile, matter should exhibit dual behaviour, that is both particle and wave like properties. However, a cricket ball of mass 100 g does not move like a wave when it is thrown by a b owler at a speed of 100km/h. calculate the wavelength of the ball and explain why it does not show wave nature. |
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Answer» Given, `m=100g=0.1kg` `v=100km//h=(100xx1000)/(60xx60)=(1000)/(36)ms^(-1)` From de-Brogile equation , wavelength `, lambda=(h)/(mv)` `lambda=(6.626xx10^(-34)kgm^(2) s^(-1))/(0.1kgxx(1000)/(36)ms^(-1))=238.5xx10^(-36)m` As the wavelength is very small so wave nature cannot be deteched. |
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| 378. |
What is the experimental evidence in support of the diea that electronic energies in an atom are quantized? |
| Answer» We know that the electrons are fast moving microparticles which have certain specific energies while revolving around the nucleus of an atom. Since electrons can lose energy only in fixed amounts or bundles called quanta and can not do so in the ground state of an atom, this means that electronic energies are quantised. | |
| 379. |
A hypothetical electromagnetic wave is shown in figure. Find out the wavelength of the radiation. |
| Answer» Wavelength `(lambda)`of radiation is the distance between two successive peaks or crusts. Therefore, `lambda=4xx(2.16"pm")=8.64 "pm"`. | |
| 380. |
Calcualte the wavelength of matter wave associated with small ball of mass of 100g travelling at a velocity of `35ms^(-1)` |
| Answer» `1.893 xx 10^(-34) m` | |
| 381. |
Out of electron and proton which one will have, a higher velocity to produce matter waves of the same wavelength ? Explain it. |
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Answer» According to de Broglie equation , `lambda=(h)/(mv)` Since both electron and proton are to have the same wavelength `(lambda)` and h is constant, therefore, `m_(e)v_(e)=m_(p)v_(p)` Now, `m_(e)` is very small as compared to `m_(p)`. Therefore, `v_(e)` will be higher as compared to `v_(p)`. |
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| 382. |
`A1`through they appear quite different to our senses, visible light, infared radiation, microwaves, radiowaves, `X` rays, and other forms of radiant enegry are all different kinds ofA. electrical radiationB. magnetic radiationC. electromagnetic radiationD. electrostatic radiation |
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Answer» Correct Answer - C Collectively, they make up the electromagnetic spectrum. |
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| 383. |
Which atom (X) is indicated by the following configuration? `Xto[Ne]3s^(2)3p^(3)`A. NitrogenB. ChlorineC. PhosphorusD. Sulphur |
| Answer» Correct Answer - C | |
| 384. |
Which of the following statements concerning the quantum numbers are correct ?A. Angular quantum number determines the three dimensional shape of the orbitalB. The pincipal quantum number determines the orientation and energy of the orbitalC. Magnetic quantum number determines the size of the orbitalD. Spin quantum number of an electron determine the orientation of the spin of electron relative to the chosen axis |
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Answer» Correct Answer - A::D (a) and (b) are the correct statements |
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| 385. |
Which of the following statements concerning the quantum numbers are correct?A. Angular quantum number determines the three dimensional shape of the orbital.B. The principal quantum number determines the orientation and energy of the orbital.C. Magnetic quantum number determines the size of the orbital.D. Spin quantum number of an electron determines the orientation of the spin of electron relative to the chosen axis. |
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Answer» Correct Answer - A::D (a,d) are both correct. The other two options are not correct. |
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| 386. |
The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying a voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal. |
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Answer» Energy of the incident radiation = Work function + Kinetic energy of photoelectron Energy of incident radiation `(E) = hv = h (c)/(lamda) = ((6.626 xx 10^(-34Js) (3.0 xx 10^(8) ms^(-1)))/((256.7 xx 10^(-9)m)))` `= 7.74 xx 10^(-19) J = 4.83 eV (1 eV = 1.602 xx 10^(-19)J)` The potential applied gives the kinetic energy to the electron. Hence, kinetic energy of the electron = 0.35 eV `:.` Work function `= 4.83 eV - 0.35 eV = 4.48 eV` |
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| 387. |
The electrons occupying the same orbital are distinguished byA. azimuthal quantum numberB. spin quantum numberC. principal quantum numberD. magnetic quantum number |
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Answer» Correct Answer - B Electrons occupying the same orbital of an atom have same value for principle quantum number, azimuthal quantum number and magnetic quantum but differ in their spin quantum number. If one has `+ (1)/(2)`, the other has `- (1)/(2)` |
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| 388. |
The ejection of the photoelectron from the silver metal in the photoelectric effect exeriment can be stopped by applying the voltage of `0.35 V` when the radiation `256.7 nm` is used. Calculate the work function for silver metal. |
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Answer» `lambda_(256.7 nm = 256.7 xx10^(-9)m"","" K.E.=0.35 eV` `E=(hc)/(lambda)=((6.626xx10^(-34)Js)(3xx10^(8) m s^(-1)))/((256.7xx10^(-19)m))` `=(6.626xx3)/(256.7)xx10^(-17)J=(6.626xx3xx10^(-17))/(256.7xx1.602xx10^(-19))eV` `=(662.6xx3)/(256.7xx1.602)eV=4.83 eV` `E=E_(0)+K.E.` `4.83 eV = E_(0)+0.35 eV` `E_(0)=4.83-0.35=4.48eV` |
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| 389. |
The ejection of the photoelectron from the silver metal in the photoelectric effect exeriment can be stopped by applying the voltage of `0.35 V` when the radiation `256.7 nm` is used. Calculate the work function for silver metal.A. `4.45 eV`B. `7.86 eV`C. `5.36 eV`D. `6.78 eV` |
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Answer» Correct Answer - A According to the phoeoelectric effect equation, `hv = hv_(0) +KE_("max")` `:. Hv_(0) (work function) = hv - KE_("max")` Find the energy of photon `(hv)` through wavelength: `hv = (hc)/(lambda) =((6.6 xx10^(-34)Js)(3.0xx10^(8)ms^(-1)))/(256.7xx10^(-9)m)` `= 0.077xx10^(-17)J = 7.7 xx 10^(-19) J` `= (7.7xx10^(-19)J) ((1eV)/(1.6xx10^(-19)J))` `= 4.8 eV` Find the maximum kinetic enegry through stopping potential: `KE_(max) = eV_(0) = 0.35eV` `:.` Work function `= (4.8eV) - (0.35eV)` `= 4.45 eV` |
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| 390. |
Table-tennis ball has mass 10g and s peed of 90m/s. if speed can be meausred within an accuracy of 4%. What will be the uncertainly in speed and position?A. `10,4xx10^(-33)`B. `10,5.27xx10^(-34)`C. `0.1,5xx10^(-34)`D. None of these |
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Answer» Correct Answer - B Uncertainty in the speed of ball `=(100xx10)/100=10m//s Deltax.mDeltav=h/(4pi)` Uncertainty in the position, `Deltax=h/(4pimDeltav)=(6.626xx10^(-34))/(4xx3.14xx10xx10^(-3)xx10)=5.27xx10^(-34)m` |
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| 391. |
Which of the following equations implies that a particle in motion can be treated as a wave and a wave can exhibit the properties of a particle?A. `2pir = n lambda`B. `lambda= (h)/(m)`C. `2pir = nh`D. `lambda = (h)/(mv)` |
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Answer» Correct Answer - D de Broglie deduced that the particle and wave properties are related by the expression `lambda = h//mv`. Note that the left side of equation involves wavelength the wave-like property, whereas the right side makes reference to mass, a distinctly particle-like priperty. |
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| 392. |
The number of d-electrons in `Fe^(2+) (Z = 26)` is not equal to the number of electrons in which one of the following ?A. d-electrons in Fe (Z = 26)B. p-electrons in Ne (Z = 10)C. s-electrons in Mg (Z = 12)D. p-electrons in Cl (Z = 17) |
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Answer» Correct Answer - D `._(26)Fe = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(6) 4s^(2)` (d - electrons = 6) `Fe^(2+) = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(6)` (d-electrons = 6) `._(10)Ne = 1s^(2) 2s^(2) 2p^(6)` (p-electrons = 6) `._(12)Mg = 1s^(2) 2s^(2) 2p^(6) 3s^(2)` (s-electron `= 2 + 2 + 2 = 6`) `._(17)Cl = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(5)` (p-electrons `= 6 + 5 = 11`) |
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| 393. |
The standing waves are generated by plucking a guitar string. The length of the string `(l)`must be equal to a whole number timesA. `lambda`B. `(lambda)/(2)`C. `(lambda)/(3)`D. `(lambda)/(4)` |
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Answer» Correct Answer - B Length of the string `=n (lambda//2)`, where `n = 1, 2, 3, ….`.The only waves that are possible are those in which a whole number of half-wavelengths fits into the string length. These allowed waves constitute a harmonic series. Any total motion of the string is some combination of these allowed harmonics. |
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| 394. |
If n = 6, the correct sequence sequence of filling of electrons will beA. `ns rarr np rarr (n - 1) d rarr (n -2) f`B. `ns rarr (n -2) f rarr (n -1) d rarr np`C. `ns rarr (n -1) d rarr (n -2) f rarr np`D. `ns rarr (n-2) f rarr np rarr (n -1) d` |
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Answer» Correct Answer - B The orbitals are 6s, 6p, 5d and 4f. Their order of filling is 6s, 4f, 5d, 6p, i.e., ns, `(n -2)f, (n-1)d, np` |
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| 395. |
When is the energy of electron regarded as zero ? |
| Answer» The energy of the electron is regarded as zero when it is at infinite from the nucleus. At that point, the forces of interaction between the electron and the nucleus are almost nil. Therefore, its energy is regarded as zero. | |
| 396. |
Choose the pair in which after filling of electron , energy sequence is reversedA. `2p gt 2s`B. `3d gt 4s`C. ` 4p lt 5s`D. `5p lt 6s` |
| Answer» Correct Answer - 2 | |
| 397. |
Chlorophyll present in green leaves of plants absorbs light at `4.620xx10^(14)Hz`. Calculate the wavelength of radiation in nanometer. Which part of the electromagnetic spectrum does it belong to? |
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Answer» For electromagnetic radiations, Wavelength )`(lambda)=(c)/(v)=((3.0xx10^(8)ms^(-1)))/((4.60xx10^(14)s^(-1)))` `=0.652xx10^(-6)m=652xx10^(-9)m.=652nm` |
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| 398. |
Given graph may belong with A. 1sB. 2pC. 3dD. All of these |
| Answer» Correct Answer - 4 | |
| 399. |
The orbital having zero probability of finding electron on the surface of nucleus isA. sB. `p_(x)`C. `d_(x^(2) -y^(2))`D. Both (2)& (3) |
| Answer» Correct Answer - 4 | |
| 400. |
The probability of finding the electron in the orbital isA. zero at the nucleusB. Maximum on two opposite sides of the nucleus along X-axisC. Zero on Z-axisD. same on all sides around the nucles. |
| Answer» Correct Answer - A::B::C | |