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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.
| 651. |
For the following statements, write T for true and F for false : (a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons (b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral (c) The mass of an electron is about `(1)/(2000)` times that of a proton (d) A radioactive isotope of iodine is used for making tincture iodine, which is used as a medicine. |
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Answer» (a) F (b) F (c) T (d) F |
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| 652. |
Which two of the following atomic species are isotopes of each other and which two are isobars ? `._(90)^(231)Z,._(91)^(230)Z,._(88)^(230)Z,._(90)^(233)Z`. |
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Answer» (a) The isotopes of an element have the same atomic number but different mass numbers. The lower figures in the above given symbols indicate the atomic numbers. Now, in this case there are two atoms having the same atomic number of 90. So, the two isotopes will be : `._(90)^(231)Z and ._(90)^(233)Z` (b) The isobars have different atomic numbers but same mass numbers. The upper figures in the given symbols indicate the mass numbers. In this case there are two atoms having the same mass number if 230. So, the two isobars will be : `._(91)^(230)Z and ._(88)^(230)Z`. |
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| 653. |
Explain with examples (i) Atomic number (ii) Mass number (iii) Isotopes, and (iv) Isobars. Give any two uses of isotopes. |
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Answer» (i) Atomic number. The number of protons in one atom of an element is known as atomic number of that element. For example, one atom of sodium element has 11 protons in it, so the atomic number of sodium is 11. Similary, one atom of carbon element has 6 protons in it, so the atomic number of carbon is 6 (ii) Mass number. The total number of protons and neutrons present in one atom of an element is known as its mass number. For example, one atom of sodium element contains 11 protons and 12 neutrons, so the mass number of sodium is `11 + 12 = 23`. Similarly, a normal carbon atom has 6 protons and 6 neutrons, so the mass number of carbon is `6+6 = 12` (iii) Isotopes. Isotopes are atoms of the same element having the same atomic number but different mass numbers. The isotopes of an element have the same atomic number because they contain the same number of protons (and electrons). The isotopes of an element of isotopes. All the chlorine atoms contain 17 protons, so the atomic number of all the chlorine atoms is 17. Now, some chlorine atoms have 18 neutrons whereas other chlorine atoms contain 20 neutrons. Chlorine atoms can, therefore, have mass numbers of `17 + 18 =35 or 17 + 20 = 37`. Thus, chlorine has two isotopes of mass numbers 35 and 37 respectively. The two isotopes of chlorine can be written as `._(17)^(35)Cl and ._(17)^(37)Cl`. (iv) Isobars. Isobars are the atoms of different elements having different atomic numbers but the same mass number. Isobars have different number of protons in their nuclei but the total number of nucleons (protons + neutrons) in them is the same. An example of isobars is argon, `._(18)^(40)Ar`, and calcium, `._(20)^(40)Ca` (The is because argon and calcium are atoms of different elements having different atomic numbers of 18 and 20 respectively but they have the same mass number of 40. Uses of Isotopes (i) Radioactive isotopes (such as Uranium-235) are used as a fuel in nuclear reactors of nuclear power plants for generating electricity (ii) Radioactive isotopes (such as Cobalt-60) are used in the treatment of cancer. |
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| 654. |
In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is `(h = 6.6 xx 10^(-34) kg m^(2) s^(-1), " mass of electron, " m_(e) = 9.1 xx 10^(-31) kg)`A. `1.52 xx 10^(-4) m`B. `5.10 xx 10^(-3) m`C. `1.92 xx 10^(-3)m`D. `3.84 xx 10^(-3) m` |
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Answer» Correct Answer - C `Delta v = (0.005)/(100) xx 600 ms^(-1) = 3 xx 10^(-2) ms^(-1)` `Delta x xx m Delta v = (h)/(4pi)` `:. Delta x = (h)/(4pi m Delta v)` `= (6.6 xx 10^(-34) kg m^(2) s^(-1))/(4 xx 3.14 xx 9.1 xx 10^(-31) kg xx 3 xx 10^(-2) ms^(-1))` `= 1.92 xx 10^(-3) m` |
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| 655. |
Cathod rays are not waves but are composed of electrically charged particles as they are deflected byA. magnetic fieldsB. electric fieldsC. both magnetic and electric fieldsD. neither magnetic nor electric fields |
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Answer» Correct Answer - C Waves cannot be deflected by magnetic and electric fields as they do not carry any charge. |
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| 656. |
The energy required to break one mole of `Cl-Cl` bonds in `Cl_2` is `242 kJ mol^-1`. The longest wavelength of light capable of breaking a since `Cl-Cl` bond isA. 494 nmB. 594 nmC. 640 nmD. 700 nm |
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Answer» Correct Answer - A Energy required to break one mole of Cl - Cl bonds `=242 kJ "mol"^(-1)` Energy required to break one Cl-Cl bond `=((242xx10^(3) "J mol"^(-1)))/((6.022xx10^(23) "mol"^(-1)))` =`4.029xx10^(19)J` We know that `E=(hc)/(lambda)` `lambda=(hc)/(E)=((6.626xx10^(-34)Js)xx3xx10^(8) ms^(-1))/((4.02xx10^(-19)J))` `=4.94xx10^(-7)m=494xx10^(-9)m=494 nm` |
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| 657. |
Statement 1. An electron can never be found in the nucleus. Statement 2. Uncertainty in the position of an electron is greater than `10^(-15)m`.A. Statement-1 is true , Statement-2 is also true. Statement-2 is the correct explanation of Statement-1B. Statement-1 is true , Statement-2 is also true. Statement-2 is not correct explanation of Statement-1C. Statement -1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
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Answer» Correct Answer - A Statement 2 is the correct explanation for Statement 1. |
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| 658. |
Statement 1. An atom is electrically neutral . Statement 2. Number of electrons is equal to number of neutrons.A. Statement-1 is true , Statement-2 is also true. Statement-2 is the correct explanation of Statement-1B. Statement-1 is true , Statement-2 is also true. Statement-2 is not correct explanation of Statement-1C. Statement -1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
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Answer» Correct Answer - C Correct statement-2. No. of electrons is equal to number of protons. |
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| 659. |
Ground state electronic configuration of Cr atom is :A. `[Ar]3d^(4)4s^(2)`B. `[Ar]3d^(5)4s^(1)`C. `[Ar]3d(6)4s^(0)`D. `[Ar]3d^(5)4s^(2)` |
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Answer» Correct Answer - B is the correct answer. |
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| 660. |
The electrons identified by quantum numbers n and l :- (a) n=4, l=1 (b) n=4, l=0 (c ) n=3, l=2 (d) n=3, l=1 Can be placed in order of increasing energy asA. `(iv) lt (ii) lt (iii) lt (i)`B. `(ii) lt (iv) lt (i) lt (iii)`C. `(iii) lt (iii) lt (ii) lt (iv)`D. `(iii) lt (i) lt (iv) lt (ii)` |
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Answer» Correct Answer - A The (n+l) values for the different electrons are : (i) 4+1=5 , (ii)4+0=4 (iii) 3+2=5 , (iv) 3+1=4 The correct increasing order of energy is : `(iv) lt (iii) lt (ii) lt(i)` |
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| 661. |
The energy of an a-particle is `6.8xx10^(-18)J` . What will be the wave length associated with it ? Given : `h=6.62xx10^(-34)Js, 1 "amu"=1.67xx10^(-27)`kg. |
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Answer» `lambda=(h)/(sqrt(2mE))" " ," " E=6.8xx10^(-18)J=6.8xx10^(-18)kgm^(2)s^(-2)` `h=6.62xx10^(-34)Js=6.62xx10^(-34)kgm^(2)s^(-1)` `m(alpha-"particle")=4"amu"=4xx1.67xx10^(-27)kg` `lambda=((6.62xx10^(-34)"kg "m^(2)s^(-1)))/([2xx(6.8xx10^(-18)"kg "^(2)s^(-2))xx(4xx1.67xx10^(-27)kg)]^(1//2))=0.22xx10^(-11)m`. |
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| 662. |
What will be the maximum number of electrons having the same spin in an atom with `n +l=4` ? |
| Answer» In an atom, both 4s (n = 4, l = 0) and (n = 3, l = 1) orbitals can have n + l =4. Since four orbitals are involved, they can have maximum of eight electrons. Out of these half of the electrons (4) can have the same spin (+1/2 or -1/2). | |
| 663. |
Match the following species with their corresponding ground state electronic configuration |
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Answer» `Ararr3Brarr4Crarr1Drarr5` (A) `C(Z=29): 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(1)` B. `cu^(2+)(Z=29):1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(9)` C. `Zn^(2+)(Z=30):1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)` D. `cr^(3+)(Z=24): 1s^(2)2s^(2)2p^(2)3s^(2)3p^(6) 3d^(3)` |
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| 664. |
Which set of quantum numbers is not possible ?A. `{:(n,l,m," s"),(3,2,0,+1//2):}`B. `{:(n,l,m," s"),(2,2,1,+1//2):}`C. `{:(n,l,m," s"),(1,0,0,-1//2):}`D. `{:(n,l,m," s"),(3,2,-2,+1//2):}` |
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Answer» Correct Answer - B `l = 0 " to " n - 1`. Hence, l cannot be equal to n |
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| 665. |
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as `v = 3.29 xx 10^(15) (Hz) [1//3^(2) - 1//n^(2)]` Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum. |
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Answer» `v = (c)/(lamda) = (3.0 xx 10^(8) ms^(-1))/(1285 xx 10^(-9) m) = 3.29 xx 10^(15) ((1)/(3^(2)) - (1)/(n^(2)))` or `(1)/(n^(2)) = (1)/(9) - (3.0 xx 10^(8) ms^(-1))/(1285 xx 10^(-9) m) xx (1)/(3.29 xx 10^(15)) = 0.111 - 0.071 = 0.04 = (1)/(25) or n^(2) = 25 or n = 5` The radiation corresponding to 1285 nm lies int eh infrared region. |
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| 666. |
Match species given in column I with the electronic configuration given in column II. |
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Answer» `Ararr4Brarr3Crarr1Drarr2` A. `Cr(Z=24)=1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(1)=[Ar]3d^(5)4s^(7)` B. `Fe^(2+)(Z=26)= 1s^(2)2 s^(2)2p^(6)3s^(2)3p^(6)3d^(6)4s^(6)=[Ar]3d^(6)4s^(0)` C. `Ni^(2+)(Z=28)= 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(6)4s^(0)=[Ar]3d^(8)4s^(0)` D. `Cu(Z=29)=1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)2s^(-1)=[Ar]3d^(10)4s^(1)` |
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| 667. |
Emission transitions in the Paschen series end at orbit `n=3` and start from orbit `n` and can be represented as `v=3.29xx10^(15) (Hz)[1//3^(2)-1//n^(2)]` Calculate the value f `n` if the transition is observed at `1285 nm`. Find the region of the spectrum. |
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Answer» `v=(3.29xx10^(15)Hz)((1)/(3^(2))-(1)/(n^(2)))` `lambda=1285 nm=1285xx10^(-19)m=1.285xx10^(-16)m` `v=(c)/(lambda)=((3xx10^(8)ms^(-1)))/((1.285xx10^(-6)m))=2.3346xx10^(14)s^(-1)` `2.3346xx10^(14)=3.29xx10^(15)[(1)/(3^(2))-(1)/(n^(2))]` `(2.3346)/(32.9)=(1)/(3^(2))-(1)/(n^(2)) or 0.071=(1)/(9)-(1)/(n^(2))` `(1)/(n^(2))=(1)/(9)-0.071=0.111-0.171=0.04` `n^(2)=(1)/(0.04)=25 or n=5` Paschen series lies in infrared of the spectrum. |
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| 668. |
Complete the following : (a) `........=1s^(2)2s^(2)sp^(6)3s^(2)3p^(3)` , (b) `X^(n+)(Z=26)=1s^(2)2s^(2)2p^(6)3s^(2)3p^(6) 3d^(5)` (c)`Fe^(2+)=1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(?)4s^(?)` , (d) `O^(2-)=1s^(2)2s^(2)2p?` |
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Answer» (a) `P = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(3)` , (b) `X^(3+)=1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(5)` (c) `Fe^(2+)=1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(6) 4s^(0)` , (d) `O^(2-) = 1s^(2) 2s^(2) 2p^(6)`. |
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| 669. |
`Na^(+)` has completely filled K and L shells. Explain. |
| Answer» A sodium ion, `Na^(+)`, has 10 electrons in it. Now, the maximum capacity of K shell is 2 electrons and that of L shell is 8 electrons. Taken together, the maximum capacity of K and L shell is `2 +8 = 10` electrons. A sodium ion `Na^(+)` has completely filled K and L shells because its 10 electrons can completely fill up K and L shells. | |
| 670. |
The extra stability of half-filled and completely-filled subshell is due toA. relatively small shieldingB. smaller coulombic repulsion enegryC. larger exchange enegryD. all of these |
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Answer» Correct Answer - D Electrons in the same subshell have equal enegry but different spatial distribution. Consequently, their shielding of one another is relative small. |
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| 671. |
What is the deviation from Aufbau Principle in case of electronic configuration of La (Z=57)? |
| Answer» The electronic configuration of La (Z=57) is : `[Xe] 5d^(1)6s^(2)`. It deviates from the Aufbau principle in the sense that the electron enters 5d orbital in preference of 4f orbital which is of lower energy. | |
| 672. |
In the grand state, an element has 13 electrons in its M-shell. The element isA. ManganeseB. ChromiumC. NickelD. Iron |
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Answer» Correct Answer - A::B `13e^(-)` in M (3rd) shell means `3s^(2) 3p^(6) 3d^(5)`. Hence complete configuration will be `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(5) 4s^(1) or 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(5) 4s^(2)` Hence, the element is chromium (Z = 24) or manganes (Z = 25) |
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| 673. |
The range of wavelength of the visible light isA. 400 - 750 nmB. 200- 400 nmC. 400 - 750 `Å`D. 200 - 400 `Å` |
| Answer» Correct Answer - 1 | |
| 674. |
The number of electrons in 3d shell for element with atomic number 26 isA. 4B. 6C. 8D. 10 |
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Answer» Correct Answer - B 3d sub-shell can have six electrons `(3d^(6))`. |
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| 675. |
Which among the following scientists was successful in accounting for the stability of atom ?A. NeilsBohrB. RutherfordC. J.J. ThomsonD. Maxwell |
| Answer» Correct Answer - 1 | |
| 676. |
Assertion(A): All isotopes of a given element show the same type of chemical behaviour. Reason(R) The chemical properties of an atom are controlled by the numb er of electron s in the atom.A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true but R is not the correct explanation of AC. A is true but R is falseD. Both A and R are false. |
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Answer» Both assertion and reason are true and reason is the correct explanantion of assertion. Isotopes have the saem atomic number i.e. same number of electrons which are responsible for their chemical behaviour. Hence, these exhibit smililar chemical properties. |
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| 677. |
According to Bohr theory, the angular momentum of an electron in 5th orbit isA. `10h//pi`B. `2.5 h//pi`C. `25 h//pi`D. `1.0 h//pi` |
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Answer» Correct Answer - B `(nh//2pi = 5h//2pi = 2.5 h//2pi)` |
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| 678. |
Bohr theory is not applicable forA. `He^(+)`B. `Li^(2+)`C. `Br^(+)`D. `H` |
| Answer» Correct Answer - 3 | |
| 679. |
The correct seuence of frequency of the electromagnetic radiations in electromagnetic spectrum isA. X-rays `gt` UV rays `gt` Microwaves ` gt` Radio wavesB. Radio waves `gt` Microwaves` gt` UV rays `gt` X -raysC. UV rays `gt` X -rays` gt` Radio waves `gt` MicrowavesD. Radiwaves `gt` Microwaves`gt` X -rays `gt` UV rays |
| Answer» Correct Answer - 1 | |
| 680. |
Assertion(A): All isotopes of a given element show the same type of chemical behaviour. Reason(R) The chemical properties of an atom are controlled by the numb er of electron s in the atom.A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true but R is not the correct explanation of A.C. A is true but R is false.D. Both A and R are false. |
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Answer» Correct Answer - A Reason is the correct explanation for assertion. |
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| 681. |
Assertion (A): All isotopes of a given element show the same type of chemical behaviour. Reason (R) : The chemical properties of an atom are controlled by the number of electrons in the atomA. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not the correct explanation of AC. A is true but R is falseD. Both A and R are false |
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Answer» Correct Answer - A All isotopes of any element have same atomic number and hence same number of protons and electrons. As electrons take part in a chemical reaction, atoms having same having same number of electrons show same type of chemical behaviour |
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| 682. |
The colour of light which has maximum frequency isA. BlueB. RedC. GreenD. Violet |
| Answer» Correct Answer - 4 | |
| 683. |
The value of `n_(1)` for Pashecn series of hydrogen spectrum is`(n_(1) =` orbit number in which electron falls)A. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - 3 | |
| 684. |
What valency will be shown by an element having atomic number 12 ? |
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Answer» To find out the valency of the element, we will have to write down its electronic configuration. Now, the atomic number of this element is 12, so its electronic configuration will be `underset("2, 8, 2")("K L M")` It has 2 electrons in its outermost shell (M shell). So, one atom of this element can lose 2 electrons to achieve the nearest inert gas electron arrangement of 2,8. Since one atom of the element loses 2 electrons to achieve the inert gas electron configuration, so its valency is 2. It is divalent. (The element having atomic number 12 is actually magnesium). |
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| 685. |
The kinetic energy of the photoelectrons depends upon theA. Intensity of striking lightB. Number of photons strikingC. Frequency of striking lightD. Number of photoelectrons ejevted |
| Answer» Correct Answer - 3 | |
| 686. |
The atomic number of an element X is 8 and that of element Y is 4. Both these elements can exhibit a valency of :A. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - B | |
| 687. |
Elements with valency 1 areA. always metalsB. always metalloidsC. either metals or non metalsD. always non metals |
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Answer» Correct Answer - c Correct answer c Element with vlalency 1 may be metals (+1 valency) or non metals (-1 valency) |
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| 688. |
Calculate the momentum of a particle which has de Broglie wavelength of `2.5xx10^(-10)m`. |
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Answer» Correct Answer - `2.65xx10^(-24) "kg " m s^(-1)` According to de Broglie equation, ` " " lambda(h)/(p) or p(h)/(lambda)` `p=((6.626xx10^(-34) "kg "m^(2)s^(-1)))/((2.5xx10^(-10)m))=2.65xx10^(-24) "kg " m s^(-1)`. |
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| 689. |
The maximum kinetic energy of the photoelectrons is related to the stopping potential `(v_(0))` thorugh the relationA. `KE_(max) = eV_(0)`B. `KE_(max) = e//V_(0)`C. `KE_(max) = eV_(0)^(2)`D. `KE_(max) = e sqrt(V_(0))` |
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Answer» Correct Answer - A Stopping potential is the negative potential applied to the anode of the photoelectric cell. As a result, the photoelectrons are repelled. They utilize a part of their `KE` to overcome the repulsion. If we go on increasing the negtive voltage of the anode, a stage will come when the entire `KE` of the photoelectrons is used up and electron become stationary. |
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| 690. |
In a sample of ethyl ethanote `(CH_(3)COOC_(2)H_(5))` the two oxygen atoms have the same number of electrons but different number of neutrons, which of the following is the correct reason for it?A. One of the oxygen atoms has gained electronsB. one of the oxygen atoms has gained two neutronsC. The two oxyen atomis are isotopesD. The two oxygen atoms are isobars |
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Answer» Correct Answer - c correct answer c The two oxygen atoms are the isotopes .These are the different atoms of the same element which have same number of electrons (or atomic number ) but different number of neutrons |
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| 691. |
Elements with valency 1 areA. always metalsB. always non-metalsC. always metalloidsD. either metals or non-metals |
| Answer» Correct Answer - D | |
| 692. |
The kinetic energy of a sub-atomic particle is `5.65xx10^(-25)J`. Calculate the frequency of the particle wave. |
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Answer» Correct Answer - `1.705xx10^(9) s^(-1)` Kinetic energy (K.E.) of the sub-atomic particle =`1//2mv^(2)` According to de Broglie relationship, `(lambda)=(h)/(m upsilon)` Frequency of the matter wave, (v) `("Velocity of matter wave")/("Wavelength of matter wave")=(upsilon)/(lambda)` Substituting the value of `lambda` in equation (ii) Frequency `(v)=(upsilon)/(h//m upsilon)=(m upsilon^(2))/(h)=(2xx(1)/(2) m upsilon^(2))/(h) =(2K.E.)/(h)` Substituting the values of kinetic energy and h , `v=((2xx5.65xx10^(-25)J))/((6.626xx10^(-34)Js))=1.705xx10^(9)s^(-1)`. |
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| 693. |
In a sample of ethyl ethanote `(CH_(3)COOC_(2)H_(5))` the two oxygen atoms have the same number of electrons but different number of neutrons, which of the following is the correct reason for it?A. one of the oxygen atoms has gained electronsB. one of the oxygen atoms has gained protonsC. the two oxygen atoms are isotopesD. the two oxygen atoms are isobars |
| Answer» Correct Answer - C | |
| 694. |
When a narrow beam while light is passed through a glass prism, the different wavelength travel thorugh the glass atA. same speedB. differebt speeds (rates)C. same intensityD. same frequency |
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Answer» Correct Answer - B Refractive index, `n = ("Speed of light in vacuum")/("Speed of light in the medium")` varies with wavelength. As a result, the white light is separated into its component colors. |
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| 695. |
Write the distribution of electrons in a carbon atom. (Atomic number of carbon = 6) |
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Answer» The atomic number of carbon is 6, so a carbon atom has 6 electrons in it. Out of 6 electrons : (i) the first 2 electrons will occupy K shelll. For this we write `underset(2)(K)` (ii) the remaining `6 - 2 = 4` electrons will go to L shell. For this we write `underset(4)(L)` So, the distribution of electrons in the carbon atom (or the electronic configuration of a carbon atom) will be : `underset("2, 4")("K L")or2,4`. |
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| 696. |
Write down the electronic configuration of an element with atomic number 14. Which group in the periodic table does this element belong to ? |
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Answer» Correct Answer - Group 14 E.C. of `._(14)X=1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(2)`. As the last shell contains 4 elemenets, group number `=10+4=14`. |
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| 697. |
Rutherfords experiments , which established the nuclear model of atom , used a beam of:-A. helium atoms, which impinged on a gold foil and got scatteredB. `beta` particles, which impinged on a gold foil and got scatteredC. helium nuclei, which impinged on a gold foil and got scatteredD. `gamma` rays, which impinged on a gold foil and got scattered |
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Answer» Correct Answer - C Projectiles used by Rutherfored were alpha particles which are high enegry, positively charged `He` ions emitted during radioactive decay. An alpha particle has charge `2+` and mass `4u`. |
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| 698. |
How many maximum spectral lines are possible if electron is present in `4^(th)` shell and only two atom are present in sample ?A. 6B. 4C. 3D. 2 |
| Answer» Correct Answer - B | |
| 699. |
STATEMENT-1 `:` `Deltax : Deltap = (h)/(4pi)` is not applicable for macroscope particles. STATEMENT-2 `:` p,orbital is an example of gerade. STATEMENT-3 `:` Due to production of matter waves, energy of the moving system decreasesA. T T TB. T T FC. T F TD. T F F |
| Answer» Correct Answer - C | |
| 700. |
All the students of class 9 were performing experiments to study the types of solutions in the science laboratory. Vikalp took some water in a beaker and heated it slowly with the help of a burner.He starded adding postassium nitrate to the hot water with a spoon and stirred it with a glass rod continuously, so that potassium nitrate goes on dissolving in water. Vikalp took the temperature of water up to `40^(@)C`, and then keeping the temperature constant, went on adding more and more of potassium nitrate to water, till no more potassium nitrate dissolved in it and some potassium nitrate is also left undissolved at the bottom of beaker. The contents of the beaker are now filtered through a filter paper arranged in a funnel. A clear solution is obtained in the form of a filtrate (a) Depending upon the amount of solute present, the solution can be classified into two groups. Name these two groups of solution (b) What type of potassium nitrate solution prepared by Vikalp at `40^(@)C` ? Define the type of solution prepared by Vikalp (c) What will happen if the potassium nitrate solution prepaerd by Vikalp at `40^(@)C` ? is heated further (say to `60^(@)C`) ? Give reason for your answer (d) What will happen if the potassium nitrate solution prepared by Vikalp at `40^(@)C` is allowed to cool (say to `20^(@)C`) ? Give reason for your answer (e) Which term/phrase can be used to convey that a maximum of 106 grams of potassium nitrate can be dissolved in 100 grams of water at a temperature of `60^(@)C` ? (f) What values are displayed by Vikalp in this episode ? |
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Answer» (a) (i) Saturated solutions, and (ii) Unsaturated solutions (b) (i) Vikap has prepared a saturated solution of potassium nitrate at `40^(@)C` (ii) A solution in which no more solute (here potassium nitrate) can be dissolved at that temperature, is called a saturated solution (c) If the saturated solution of potassium nitrate at `40^(@)C` is heated further to a higher temperature, then it will become an unsaturated solution (and more of potassium nitrate can then be dissolved in it). This is because the solubility of potassium nitrate in water increases on heating (or raising the temperature) (d) If the saturated solution potassium nitrate at `40^(@)C` is allowed to cool, then some of the dissolve potassium nitrate from the solution will separate out as a solid and settle down at the bottom of the beaker. This is because the solubility of potassium nitrate in water decreases on cooling (or lowering the temperature) (e) The solubility of potassium nitrate in water is 106g at `60^(@)C` (f) The values displayed by Vikalp in this episode are (i) Awareness of saturated and saturated solutions (ii) Knowledge of effect of heating and cooling on saturated solutions, and (iii) Applications of knowledge in solving problems. |
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