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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.
151. |
What is the value of `m` in the Rydberg formula for the third spectral line of the Lyman series?A. `0`B. `1`C. `4`D. `3` |
Answer» Correct Answer - C `n = 1, m = 2, 3, 4,…..` |
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152. |
The Vividh Bharti station of All India Radio , Delhi, broadcasts on a frequency of1,368 kHz. Calculate the wavelength of the electromagnetic spectrum it belongs to Hint `: lambda= ( c )/( v) ` `1kHz =10^(3)Hz` |
Answer» Frequency `(v ) = 1368kHz` `= 1368 xx 10^(3) Hz ( s^(-1))` `( 1 kH z = 10^(3) Hz)` `lambda= ( c )/( v) = ( 3.0 xx 10^(8)ms^(-1))/( 1368 xx 10^(3) s^(-1))` `= 219.3 m` This belong to radio wave wavelength |
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153. |
Photoelectric emission is observed from a surface for frequencies `v_(1) and v_(2)` of the incident radiation `(v_(1) gtt v_(2))`. If the maximum kinetic energies of the photoelectrons in the two cases are in the ratio `1 : k`, what will be the theshold frequency `(v_(0))` in term of `v_(1), v_(2)` and k ? |
Answer» In 1st case, K.E. `= h v_(1) - h v_(0) = h (v_(1) - v_(0))` In 2nd case, K.E. `= h v_(2) - h v_(0) = h (v_(2) - v_(0))` Given `(h (v_(1) - v_(0)))/(h(v_(2) - v_(0))) = (1)/(k) or v_(2) -v_(0) = k (v_(1) - v_(0)) = k v_(1) - kv_(0)` or `k v_(0) - v_(0) = kv_(1) - v_(2) or v_(0) (k -1) = kv_(1) = v_(2) or v_(0) = (kv_(1) - v_(2))/(k -1)` |
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154. |
Photoelectric emmision is observed from a surface when lights of frequency `n_(1)`and `n_(2)` incident.If the ratio of maximum kinetic energy in two cases is `K: 1` then ( Assume `n_(1) gt n_(2))` threshod frequency isA. `(K-1)xx(Kn_(2) -n_(1))`B. `(Kn_(1)-n_(2))/(1-K)`C. `(K-1)/(Kn_(1)-n_(2))`D. `(Kn_(2)-n_(1))/(K-1)` |
Answer» Correct Answer - 4 | |
155. |
A student takes some water in a beaker and heats it over a flame for determining its boiling point. He keeps on taking its temperature reading. He would observe that the temperature of water :A. keeps on increasing regularlyB. keeps on increasing ireegularlyC. first increases slowly, then decreases rapidly and eventually becomes constantD. first increases gradually and then becomes constant |
Answer» Correct Answer - D | |
156. |
Four students (A), (B), (C) and (D) independently observed the evaporation of water under different conditions, and recorded the temperature of water at regular interval as shown below The correct recording of observations is that of the student :A. AB. BC. CD. D |
Answer» Correct Answer - B | |
157. |
A student takes a mixture of sand and ammonium chloride in a china dish and heats it under a funnel fitted with a plug over a flame. He would observe that :A. solid sand gets deposited on the lower cooler parts of the funnel while solid ammonium chloride remains in the china dishB. sand and ammonium chloride get deposited on hotter parts of the funnelC. ammonium chloride gets deposited on the cooler parts of the funnel and sand remains in the china dish.D. sand collects on cooler parts of the funnel while ammonium chloride melts in the china dish |
Answer» Correct Answer - C | |
158. |
What is the ratio of the velocities of `CH_(4)` and `O_(2)` molecules such that they are associated with de broglie waves of equal wavelength? |
Answer» According to de Broglie equation,`lambda=h//m upsilon` For methane `(CH_(4)), " " lambda_(CH_(4))=(h)/(m_(CH_(4))xx upsilon_(CH_(4)))` For oxygen `(O_(2)), " " lambda_(O_(2))=(h)/(m_(O_(2))xxupsilon_(O_(2)))` Since the wavelengths of `CH_(4)` and `O_(2)` are to be equal, `:. " " lambda_(CH_(4))=lambda_(O_(2)) " " "or" (h)/(m_(CH_(4))xx upsilon_(CH_(4)))=(h)/(m_(O_(2))xxupsilon_(O_(2)))` `m_(CH_(4))xx upsilon_(CH_(4))=m_(O_(2))xxupsilon_(O_(2)) " " "or" (upsilon _(CH_(4)))/(upsilon_(O_(2)))=(m_(O_(2)))/(m_(CH_(4)))=(32)/(16)=2`. |
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159. |
In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of `3.15xx10^(-18)J` from the radiations of `600 nm`, calculate the number of photons received by the detector. |
Answer» Where,`lambda =` wavelength of radiation h = Planck’s constant c = velocity of radiation Substituting the values in the given expression of E: `E = ((6.626 xx 10^(-34) Js)(3 xx 10^(8) ms^(-1)))/((600 xx 10^(-9) m))` `E = 3.313 xx 10^(-19) J` Energy of one photon `= 3.313 xx 10^(-19) J` Number of photons received with `3.15 × 10^(-18) J` energy `= (3.15 xx 10^(-18)J)/(3.313 xx 10^(-19) J)` `= 9.5` `~~ 10` |
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160. |
Neon gas is generally used in the sign boards. If it emits strongly at `616 nm`, calculate a. The frequency of emission, b. The distance traveled by this radiation in `30 s` c. The energy of quantum and d. The number of quanta present if it produces `2 J` of energy. |
Answer» Wavelength of radiation emitted = 616 nm` = 616 × 10^(-9) m` (Given) (a) Frequency of emission (v) `v = (c)./(lambda)` c = velocity of radiation `lambda =` wavelength of radiation Substituting the values in the given expression of (v) : `v = (3.0 xx 10^(8) m//s)/(616 xx 10^(-9) m)` `= 4.87 xx 10^(8) xx 10^(9) xx 10^(-3) s^(-1)` `v = 4.87 xx 10^(14) s^(-1)` Frequency of emission `(v) = 4.87 xx 10^(14) s^(-1)` (b) Velocity of radiation, `(c). = 3.0 × 10^(8) ms^(1)` Distance travelled by this radiation in 30 s `= (3.0 xx 10^(8)ms^(-1))(30s)` `= 9.0 xx 10^(9) m` (c). Energy of quantum `(E) = hν` `(6.626 xx 10^(-34)Js)(4.87 xx 10^(14)s^(-1))` Energy of quantum `(E) = 32.87 xx 10^(-20) J` (d) Energy of one photon (quantum) `= 32.27 xx 10^(-20) J` Therefore, `32.27 xx 10^(-20) J` of energy is present in 1 quantum. Number of quanta in 2 J of energy `= (2J)/(32.27 xx 10^(-20) J)` `= 6.19 xx 10^(18)` `= 6.2 xx 10^(18)` |
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161. |
Using the Aufban principal , write the electron configuration for the gropuped srtate of the following atomic boron `(Z = 5)` neon `(Z = 10)` aluminum `(Z = 13)` chlorine `(Z = 17)` calcium `(Z = 20)` , rabidium `(Z = 13)` |
Answer» (i) Boron `(Z=5) , 1s^(2)2s^(2)2p^(1)` (ii) Neon `(Z=10) , 1s^(2) 2s^(2) 2p^(6)` (iii) Aluminium `(Z=13) , 1s^(2)2s^(2)2p^(6)3s^(2)3p^(1)` (iv) Chorine `(Z=17) , 1s^(2) 2s^(2)2p^(6)3s^(2)3p^(5)` (v) Calcium ` (Z=20) , 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6) 4s^(2)` (vi) Rubidium ` (Z=37) , 1s^(2) 2s^(2)2p^(6)3s^(2)3p^(6)3d^(10) 4s^(2) 4p^(6) 5s^(1)`. |
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162. |
If uncertainties in the measurement of position and momentum are equal, then uncertainty in the measurement of velocity is |
Answer» According to uncertainty principle, `Deltax.Deltap=(h)/(4pi)or(Deltap)^(2)=(h)/(4pi) " "`(Given `Deltax=Deltap`) `Deltap=sqrt((h)/(4pi)) or m Deltaupsilon=sqrt((h)/(4pi)) , Delta upsilon=(1)/(2m) sqrt((h)/(pi))`. |
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163. |
The radus of hydrogen atom in the ground state `0.53 Å`. The radius of `Li^(2+)` ion (Atomic number = 3) in a similar state isA. 1.06 ÅB. 0.265 ÅC. 0.17 ÅD. 0.53 Å |
Answer» Correct Answer - C Radius of `n^(th)` orbit is given by `r_(n)=(r_(0)xxn^(2))/Z` For `._(3)Li^(2+),r=r_(0)/3=0.53/3=0.176Å` |
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164. |
If the ionization energy for the hydrogen atom is 13.6 eV , the energy required to excite it from the ground state to the next higher state is nearlyA. 3.4 eVB. 10.2 eVC. 17.2 eVD. 13.6 eV |
Answer» Correct Answer - A `E_(n)=(-E_(1))/(n^(2)), E_(1)=13.6, E_(n_2)=13.6/4=3.4eV` |
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165. |
`A` sodium street light gives off yellow light that has a wavelength of `589nm`. What is the frequency of this light? |
Answer» `v=(c)/(lambda)=((3xx10^(8)ms^(-1)))/((589 xx 10^(-9)m))=5.093 xx 10^(14)s^(-1)` | |
166. |
Wavelengths of different radiations are given below: (A) 300nm (B) `300 mu m` (C) 3nm (`30A^(@)` Arrange these radiations in the increasing order of their energies. |
Answer» Let us convert the different radiations in the same units i.e. metre (m) `lambda(A) =300 nm = 300 xx10^(-9)=3xx10^(-7)m` `lambda (B) = 300 mu m = 300xx10^(-6) = 3xx10^(-4)m` `lambda(C)=3 nm = 3xx10^(-9) = 3xx10^(-9)m` `lambda(D) = 30 Ã… = 30 xx 10^(-10)=3xx10^(-9)m`. Energy (E) and wavelength `(lambda)` are related to each other as E `pro 1//lambda`. Thus , greater the wavelength of a radiation, lesser is its energy. ln the light of this, the increasing order of energies is : `B lt A lt C=D` |
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167. |
The energines `E_(1) and E_(2)` of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths, i.e., `lamda_(1) and lamda_(2)` will beA. `lamda_(1) = (1)/(2) lamda_(2)`B. `lamda_(1) = lamda_(2)`C. `lamda_(1) = 2 lamda_(2)`D. `lamda_(1) = 4 lamda_(2)` |
Answer» Correct Answer - C `E = hv = h (c)/(lamda) :. (E_(1))/(E_(2)) = (lamda_(2))/(lamda_(1))` i.e., `(lamda_(1))/(lamda_(2)) = (E_(2))/(E_(1)) = (50eV)/(25 eV) = 2 or lamda_(1) = 2 lamda_(2)` |
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168. |
The light-blue glow given off by mercury street-lamps has a wavelength of `436 nm`. What is its frequency in hertz? Strategy: We given a wavelength and need to find the frequency. Wavelength and frequency are inversely related by the equation `lambdav = c`, which can be solved for `v`. Remember to convert `lambda` from nanometers to meters. |
Answer» According to Eq. we have `v = (c )/(lambda) = ((3.00xx10^(8)(m)/(s)))/((436nm)((10^(-9)m)/(1nm)))` `6.88xx10^(14)s^(-1)` `= 6.88 xx 10^(14) Hz` |
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169. |
Assertion : X-rays are used to study the interior of objects . Reason : X-rays have very short wavelengths and possess electromagnetic character.A. if both assertion and reason are correct and reason is correct explanation for assertion .B. If both assertion and reason are correct but reason in not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect. |
Answer» Correct Answer - B Correct explanation. Since X-rays are not deflected by electric and magnetic fields and have also very high penetrating power through the matter, these can be used to study interior of objects. |
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170. |
Each type of electromagnetic radiation is spread over a specific range of wavelengths (and frequencies). The visible region ranges fromA. `500nm` to `800 nm`B. `400 nm` to `750 nm`C. `400 nm` to `700 nm`D. `500 nm` to `850 nm` |
Answer» Correct Answer - C It ranges from violet `(400 nm)` to red `(700nm)`. |
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171. |
Electromagnetic radiation with maximum wavelengths is :A. ultra violetB. radio waveC. X-rayD. infra-red |
Answer» Correct Answer - B Radiowaves have minimum frequency and maximum wavelength. |
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172. |
Different kinds of electromagnetic radiations are simply electromagnetic waves with different (i) wavelengths (ii) frequencies (iii) speedsA. `(i),(ii),(iii)`B. `(i),(iii)`C. `(ii),(iii)`D. `(i),(ii)` |
Answer» Correct Answer - D Electromagnetic waves travle at a speed of `3.00 xx 10^(8)m s^(-1)` (rounded off) or `186,000 `miles per seond, which is the speed of light in vacuum. This speed does differ from one medium to another, but the variations are insignificant for our calculations. By convention, we use the symbol `c` for the speed of electromagnetic waves, or as it is more commonly called, the speed of light. |
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173. |
The electromagnetic spectrum consists of a continuous range of wavelengths and frequencies, form_________at the lowest frequency end to_____at the highest frequency end.A. radio waves, gamma raysB. microwaves, `X` raysC. microwaves, gamma raysD. radio waves, `X` rays |
Answer» Correct Answer - A Radio waves have the longest wavelength while gamma rays have the shortest wavelength. |
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174. |
STATEMENT-1 `:` The energy of an electron depends only upon principal quantum number in case of hydrogen and hydrogen like ions. and STATEMENT -2 `:` The energy of an electron depends on principal quantum number as well as azimuthal quantum number for other atoms.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 isTrue, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False , Statement-2 isTrue |
Answer» Correct Answer - 2 | |
175. |
The pair of ions having same electronic configuration is__________.A. `Cr^(3+),Fe^(3+)`B. `Fe^(3+),Mn^(2+)`C. `Fe^(3+),Co^(3+)`D. `Sc^(3+),Cr^(3+)` |
Answer» Correct Answer - B Cr(24) : `3d^(5)4s^(1),Cr^(3+):3d^(3)` Fe(26) : `3d^(6)4s^(2),Fe^(3+):3d^(5)` Mn(25) : `3d^(5)4s^(2),Mn^(2+):3d^(5)` Co(27) : `3d^(7)4s^(2),Co^(3+):3d^(6)` Sc(21) : `3d^(1)4s^(2),Sc^(3+):3d^(0)` Thus, `Fe^(3+) and Mn^(2+)` have same electronic configuration. |
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176. |
The work function for sodium metal is `2.46eV`. Determine the cutoff wavelength for sodium? Strategy: The cutoff frequency `v_(0)` is related to the work function through the relation `v_(0) = W_(0)//h`. This corresponds to a cutoff wavelength of `lambda_(0) = (C )/(v_(0)) = (C)/(W_(0)//h) = (hc)/(W_(0))`. wavelength grater than `lambda_(0)` for a metal with work function `W_(0)` produce no photoelectric effect. |
Answer» Step `1.` First convert work function `W_(0)` from electron volts to joules. `W_(0) = 2.46eV` `=(2.46eV) ((1.6xx10^(-19)J)/(1eV))` `= 3.9 xx 10^(-19)J` Step `2.` Now solve the equation for the cutoff wavelength `lambda_(0) = (hc)/(W_(0))` `=((6.6xx10^(-34)Js)(3.0xx10^(8)ms^(-1)))/((3.9xx10^(-19)J))` `= 5.1 xx 10^(-7)m` `= 510 nm (1nm = 10^(-9)m)` |
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177. |
Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.A. `3.0 BM`B. `4.9 BM`C. `5.9BM`D. `6.9 BM` |
Answer» Correct Answer - C Atomic number `25` corresponds to the first transition series. For these elements, the magnetic moment associted with unpaired electron is given as `mu = sqrt(n(n +1))` where `n` is the number of unpaired electrons. `(Z = 25)rArr 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)4s^(2)`. After the removal of `2e^(-)s`, it become `[Ar] 3d^(5)`having unpiared eletrons. thus, `mu = sqrt(5(5+2))BM = 5.9 BM` |
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178. |
Which of the following statement is not correct about the characterstics of cathode raysA. They start from the cathode and move towards the anode.B. They travel in straight line in the absence of an external electrical or magnetic field,C. Characteristics of cathode rays do not depend upon the material of electrodes in cathode ray tube.D. Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube. |
Answer» Correct Answer - D The characteristics of cathode rays (electrons) do not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube. |
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179. |
Assertion : X-rays are used to study the interior of objects . Reason : X-rays have very short wavelengths and possess electromagnetic character.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
Answer» Correct Answer - B X-rays are used to study the interior of the objects because they are not deflected by the electric and magnetic field and have a very high penetrating power through the matter. |
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180. |
The number of d-electron retained in `Fe^(2) ("At no. of" Fe =26)` ion is.A. `6`B. `5`C. `4`D. `3` |
Answer» Correct Answer - A `(Z = 26): 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(6)4s^(2)` or simply `[Ar] 3d^(6)4s^(2)`. Thus, the ground state electron configuration of `Fe^(2+)` will be `[Ar]3d^(6)`. Therfore, it retains `6d` electrons. Never write the configuration as `[Ar]4s^(2)3d^(6)` because we many make the mistake of ionizing `3d`electrons first leaving behind just four `d` electrons. |
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181. |
Which of the following ions has the maximum magnetic moment in aqueous solution ?A. `Fe^(2+)`B. `Ti^(2+)`C. `Mn^(2+)`D. `Cr^(2+)` |
Answer» Correct Answer - C The magnitude of magnetic moment is directly related to the number of unpaired electrons: `Ti(Z = 22):[Ar]3d^(2)4s^(2)`, thus `Ti^(2+)`has `2` unpaired electorns. `Cr(Z = 24):[Ar]3d^(5)4s^(1)`, thus `Cr^(2+)`has `4` unpaired electrons. `Mn(Z = 25): [Ar]3d^(5)4s^(2)`, thus `Mn^(2+)`has `5` unpaired electrons. `Fe(Z = 26):[Ar]3d^(6)4s^(2)`, thus `Fe^(2+)`has `4` unpaired electorns. |
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182. |
The absolute charge on an electron is -A. `-1.6 xx 10^(-19) C`B. `9.1 xx 10^(-31)C`C. ` - 1.6 xx 10^(-31) C`D. ` - 9.1 xx 10^(-19)C` |
Answer» Correct Answer - 1 | |
183. |
Thomson found that cathode rays move at about__________the speed of light.A. one-fifthB. one-sixthC. one-fourthD. one-third |
Answer» Correct Answer - A Speed of light is `30 xx 10^(8) ms^(-1)`. |
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184. |
Which of the following statement is not correct about the characteristics of cathode rays ?A. They start from the cathode and move towards the anodeB. They travel in straight line in the absence of an external electrical or magnetic fieldC. Characteristics of cathode rays do not depend upon the material of electrodes in cathode ray tubeD. Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube |
Answer» Correct Answer - D Characteristics of cathode rays do not depend upon the nature of the gas in cathode ray tube |
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185. |
The threshold frequency, `v_(0)`, of a metal is `6.7xx10^(14)s^(-1)`. Calculate the maximum kinetic enegry of a single electron that is emitted when a radiation of frequency `v = 1.0 xx 10^(-15)s^(-1)` strikes the metal. Strategy: Use the relatiship between `v_(0),v,`and `KE` given in Eq. |
Answer» Step `I.` The relatiship between `v_(0),v`, aand maximum kinetic enegry is given as `KE_(max) = hv - hv_(0)` Step `2`. Solving the equation for kinetic energy gives `KE_(max) = h(v - v_(0))` `=(6.6xx10^(-34)Js)(1.0 xx 10^(15)s^(-1)-6.7xx10^(14)s^(-1))` `=(6.6xx10^(-34) Js) (10xx10^(14)s^(-1) - 6.7 xx 10^(14) s^(-1))` `=(6.6 xx 10^(-34)Js) (3.3xx10^(14)s^(-1))` `= 2.2xx10^(-19)J` |
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186. |
Cathode rays move towardsA. Positively charged cathodeB. Positively charged anodeC. Negatively charged cathodeD. Negatively charged anode |
Answer» Correct Answer - 2 | |
187. |
Assertion : The characteristics of cathode rays do not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube. Reason : Cathode rays consist of negatively charged particles, called electrons.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
Answer» Correct Answer - A Cathode rays consist of electrons which are the basic constituent of all the atom and hence their characteristics do not depend upon the material of electrodes and nature of the gas present in the cathode ray tube. |
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188. |
A strong argument in favour of the particle nature of cathode rays is that theyA. produce fluorescenceB. travel through vacuumC. can rotate a light paddle wheel placed in their pathD. cast shadow of the objects lying in their path |
Answer» Correct Answer - C | |
189. |
Why the cathode rays always are considered to be consiste of fundamental common particles called electrons? |
Answer» This is because the electrons possess same `( e )/( m) `ratio irrespective of the gas in the discharge tube or the nature of electrodes .This means that the cathode rays do not consist of gaseous charged atoms,otherwise `(e )/( m)` ratio would have been different for different gases taken. | |
190. |
An isotone of `._(32)^(76)Ge` is-(a) `._(32)^(77)Ge`(b)`._(33)^(77)As`(c)`._(34)^(77)Se`(d)`._(34)^(78)Se`A. `._(32)^(77)Ge`B. `._(33)^(77)As`C. `._(34)^(77)Se`D. `._(34)^(78)Se` |
Answer» Correct Answer - B::D `._(33)^(77)As` and `._(34)^(78)Se` have same number of neutrons `=(A-Z) "as" ._(32)^(56)Ge`. i.e.,44. |
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191. |
The wavelength of a violet radiation is `3.7xx10^(5)` pm. What is its frequency ? |
Answer» Correct Answer - `8.1xx10^(14) s^(-1)` `c=3xx10^(8) m s^(-1), lambda=3.7xx10^(5)xx10^(-12)=3.7xx10^(-7)m`. `v=(c)/(lambda)=((3xx10^(8)ms^(-1)))/((3.7xx10^(-7)m))=8.1xx10^(14)s^(-1)` |
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192. |
What is the frequency of light with a wavelength of 480 nm |
Answer» Correct Answer - `v=6.25xx10^(14)s^(-1), bar(v)=2.08xx10^(6)m` `lambda=480nm=480 xx10^(-9)m, c=3xx10^(8)m^(-1)` `v=(c)/(lambda)=((3xx10^(8)ms^(-1)))/((480xx10^(-9)m))=(3)/(48)xx10^(16)s^(-1)=6.25xx10^(14)s^(-1)` Wave number `(bar(v))=(1)/(lambda)=(1)/(480xx10^(-9)m)=2.08xx10^(6)m` |
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193. |
Assertion : The electronic configuration of K) (19) is `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(1)`. Reason : Energy of `4s lt 3d`. This is decided by Aufbau principle.A. if both assertion and reason are correct and reason is correct explanation for assertion .B. If both assertion and reason are correct but reason in not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect. |
Answer» Correct Answer - A Reason is the correct explanation for assertion. |
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194. |
Calculate the number of protons , electrons and neutrons in the following. (a) Chloride ion `(Cl^(-))` with `Z = 17,A =35` (b) Aluminium ions`(Al^(3+)) ` with `Z = 13 , A=27` |
Answer» (a)`Cl^(-) ` with `Z = 17 , A= 35` Number of protons`= 17` Number of electrons `=17 +1` ( As one negative charge which means gains of electron) `=18` Number of neutrons `= A-Z = 35 - 17 = 18` ( b) `Al^(3+)` with `Z= 13 ` `A=27` Number of protons `= 13` Number of electrons `=13-3 = 10 ` ( As it has three positive charge obtained by the loss of three electrons) number of neutrons `= A-Z= 27-13=14` |
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195. |
Why is charge on cathode ray particles ? |
Answer» (a) The cathode rays move from cathode to anode. (b) They travel in straight lines. (c ) They rotate little paddlewheelin their path. (d) They attract towards `( +)` ve electrode of electric field. |
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196. |
What additional information is needed to answer the following: Which ion has the electronic configuration : `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)`? |
Answer» In order to identity the ion, the charge on the ion is required. | |
197. |
An isotone of `._(32)^(76)Ge` is (i) `._(33)^(77)As` (ii) `._(34)^(77)Se` (iii) `._(34)^(78)Se` (iv) `._(32)^(77)Ge`A. `(i),(iii)`B. `(ii),(iv)`C. `(i),(iv)`D. `(iii),(iv)` |
Answer» Correct Answer - A Isotones are the atoms of different elements containing the same number of neutrons. They differ in atomic number as well as mass number. Number of neutrons `(N)` is equal to mass number `(A)` minus atomic number `(Z)`. `._(32)^(76)Ge (N = 44), _(33)^(77)As(N = 44), _(34)^(77)Se(N = 43)` `._(34)^(78)Se (N = 44), _(32)^(77)Ge (N = 45)` |
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198. |
Name the dipositive ion represented by configuration : `1s^(2) 2s^(2) 2p^(6)3s^(2)3p^(6)`. |
Answer» The dipositive ion is `Ca^(2+)` ion. The configuration of Ca is `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)`. | |
199. |
Name a dipositive metal ion with configuration : `1s^(2) 2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)` |
Answer» The configuration of ion `M^(2+) : 1s^(2)2s^(2)*2p^(6)3s^(2)3p^(6)3d^(5)`. `" " darr + 2e^(-)` The configuration of atom (M) `: 1s^(2)*2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(5)` The atom represent manganese (Mn) and the dipositive ion is `Mn^(2+)ion`. |
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200. |
Find the total number of electrons in a molecule of phosphoric acid `(H_(3)PO_(4))`. |
Answer» Atomic No. of P=15 , O=8 and H=1 A molecule of `H_(3)PO_(4)=3 H`atoms + 1 P atom + 4 O atoms `:.` Total no of electrons present `=3xx1+1xx15+4xx8=50`. |
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