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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.
| 1101. |
Which of the following is the energy of a possible excited state of hydrogen?A. `3.4eV`B. `+6.8eV`C. `+13.6eV`D. `-6.8eV` |
| Answer» Correct Answer - A | |
| 1102. |
For the energy level is an atom which one of the following statement is correct ?A. There are seven principal electron energy levelsB. The second principal energy level can have four sub-energy levels and contains a maximum of eight electronsC. The M energy level can have maximum of 32 electronsD. The 4s sub-energy level is at a higher energy than the 3d sub-energy level |
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Answer» Correct Answer - B The second principal shell contains four orbitals viz 2s ,`2p_x, 2p_y` and `2p_z` |
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| 1103. |
To what effective potential a proton beam be subjected to give its protons a wavelength of `1xx10^(-10)m`? |
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Answer» Correct Answer - `0.0826 `volts `V=((h)/(lambda))_(2)^(2)(1)/(mp.e)` `=8.26xx10^(-2) V` |
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| 1104. |
In a H-like species there are two energy levels A and B above the ground state having principal quantum numbers of `n_(1) "and"n_(2)` respectively. A sample of this H-like species has all atoms`//`ions in excited levels A or B only and none in any other energy level. Energy of level B is greater than that of level A and a total of 15 different lines are emitted from this sample on returning to ground state out of which 6 lines are emitted due to electronic transitions between the level `n_(1)"and"n_(2)` only. Also energy difference between levels `n_(2) "and"n_(1), E_(n_2)-E_(n_2)=4.53 eV` The correct option is :A. `n_(2)=2,n_(1)=5`B. `n_(2)=6,n_(1)=3`C. `n_(2)=5,n_(1)=2`D. `n_(2)=3,n_(1)=6` |
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Answer» Correct Answer - b |
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| 1105. |
In a H-like species there are two energy levels A and B above the ground state having principal quantum numbers of `n_(1) "and"n_(2)` respectively. A sample of this H-like species has all atoms`//`ions in excited levels A or B only and none in any other energy level. Energy of level B is greater than that of level A and a total of 15 different lines are emitted from this sample on returning to ground state out of which 6 lines are emitted due to electronic transitions between the level `n_(1)"and"n_(2)` only. Also energy difference between levels `n_(2) "and"n_(1), E_(n_2)-E_(n_2)=4.53 eV` Calculate minimum wavelength of the above transition:A. `(4)/(3R_(H))`B. `(9)/(8R_(H))`C. `(900)/(11R_(H))`D. `(36)/(35R_(H))` |
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Answer» Correct Answer - d |
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| 1106. |
Suppose that a hypothetical atom gives a red, green, blue and violet line spectrum. Which jump according to figure would give off the red spectral line. .A. `3 to 1`B. `2 to 1`C. `4 to 1`D. `3 to 2` |
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Answer» Correct Answer - D |
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| 1107. |
Suppose that a hypothetical atom gives a red, green, blue and violet line spectrum. Which jump according to figure would give off the red spectral line. .A. `3 rarr 1`B. `2 rarr 1`C. `4 rarr 1`D. `3 rarr 2` |
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Answer» Correct Answer - D (d) According to energy, `E_(4 rarr1) gt E_(3 rarr 1) gt E_(2 rarr 1) gt E_(3 rarr 2)`. According to energy, Violet gt Blue gt Green gt Red. :. `"Red line" rArr 3 rarr 2 "transtition"`. |
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| 1108. |
Calculate the energy of photons of radiation whose wavelength is `5000A^(@)`? |
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Answer» `lambda = 5000 A^(@) = 5000 xx 10^(-10)m = 5 xx 10^(-7)m` `c = 3 xx 10^(8)m//sec, h = 6.62 xx 10^(-34)js` `E = (hc)/(lambda) = (6.62 xx 10^(-34) xx 3xx 10^(8))/(5 xx 10^(-7))` `= 3.97 xx 10^(-19)J` |
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| 1109. |
How many sub-shell are there in N shell? How many orbitals are there in d sub-shell? |
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Answer» Correct Answer - For `N` shell, principal quantum number, `n=4` Number of sub-shell, in an energy level `=n` `therefore` Number of sub-shells, in `N` shell `=4` For `d` subshell `I=2` Number of orbitals in a sub-shell `=2l+1` Number of orbitals in a d-sub-shell `=2xx2+1=5` |
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| 1110. |
The wave number of the first line of Balmer series of hydrogen is `15200 cm ^(-1)` The wave number of the first Balmer line of `Li^(2+)` ion isA. `2.4 xx 10^(5) cm^(-1)`B. `24.3 xx 10^(5) cm^(-1)`C. `6.08 xx 10^(5) cm^(-1)`D. `60.8 xx 10^(5) cm^(-1)` |
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Answer» Correct Answer - A |
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| 1111. |
A hydrogen like species is in a spherical symmetrical orbital `S_(1)` having 3 radial nodes. It gets de-excited to another level `S_(2)` having no radial node. Energy of `S_(2)` orbital is 2.25 times energy of 1st Bohr robit of hydrogen atom. ltbr. What is the combined total number of nodes `("radial" +"angular) is" S_(1) and S_(2)?`A. 4B. 3C. 5D. 6 |
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Answer» Correct Answer - a |
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| 1112. |
For the Hydrogen spectrum , last line of the Lyman series has frequency `v_(1)` , last line of Lyman series of `He^(+)` ion has frequency `v_(2)` and last line of Balmer series ofA. `2 (v_(1) + v_(3)) = v_(2)`B. `v_(1) = v_(3)`C. `4v_(1) = v_(2)`D. All of these |
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Answer» Correct Answer - D |
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| 1113. |
If the wavelength of green light is about `5000A^(@)`, then the frequency of its wave isA. `16 xx 10^(14)sec^(-1)`B. `16 xx 10^(-14)sec^(-1)`C. `6 xx 10^(14)sec^(-1)`D. `6 xx 10^(14)sec^(-1)` |
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Answer» Correct Answer - C `v = (c)/(lambda)` |
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| 1114. |
The energy of photon of light having frequency of `3 xx 10^(15)S^(-1)` isA. `1.99xx 10^(-18)J`B. `1.99 xx 10^(-17)J`C. `1.99 xx 10^(-17)ergs`D. `1.99 xx 10^(-18) ergs` |
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Answer» Correct Answer - A `E = hv` |
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| 1115. |
`A` wave function for an atomic orbital is given as `Psi_(2,1,0)` Recogine the orbital |
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Answer» Correct Answer - `Psi_(n,l,m)=Psi_(2,1,0)` `n=2,l=1,m=0` `{: (l=1,p"subshell"),(n=2,L"shell"),(m=0,p_(z)"orbital"):}` |
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| 1116. |
The sub-energy level having minimum energy isA. 3dB. 5pC. 4sD. 4p |
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Answer» Correct Answer - C Lower `(n+l)` indicates lower energy |
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| 1117. |
Electron falls from `7^"th"` energy level and lower energy levels to produce bands in the Paschen series. The total number of bands obtained will be |
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Answer» Correct Answer - 4 Paschen series is produced by jump to n=3 and starting n=7, so possible band produced for transition are `7to3, 6to3, 5to 3` and `4 to 3` |
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| 1118. |
`{:(Column I,Column II),((A)"Radial function" psi_((r)),(p)"Principle" Q.No .),((B)"Angular function"psi_((theta)),(q)"Azimuthal"Q.No .),((C)"Angular function" psi_((phi)),(r)"Magnetic" Q.No),((D)"Quantized angular momentum",(s)"Spin" Q.No),(,(t)"Shape of orbital"):}` |
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Answer» Correct Answer - A-p,q; B-q,t; C-r,t; D-q,s Based upon orientation of electrons |
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| 1119. |
When the azimuthal quantum number l = 1, the shape of the orbital will beA. UnsymmetricalB. Spherically symmetricalC. Dumb-bellD. Complicated |
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Answer» Correct Answer - C For p-orbital , l=1 means dumb-bell shape |
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| 1120. |
The orbitals amongst the following , having three nodal surfaces:A. 1sB. 2sC. 3sD. 4s |
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Answer» Correct Answer - B |
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| 1121. |
The electronic configuration of an element is `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)4s^(1)` .This represents itsA. excited stateB. ground stateC. cationic formD. anionic from |
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Answer» Correct Answer - B (b) `3 d^5 4 s^1` system is more stable than `3 d^4 4 s^2`, hence former is the ground state configuration. |
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| 1122. |
The quantum numbers `n = 2, 1 = 1` represent.A. `1 s` obitalB. `2 s` orbitalC. `2 p` orbitalD. `3 d` orbital |
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Answer» Correct Answer - C ( c) `l = 1` is for `p` orbital. |
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| 1123. |
The number of d-electrons retained in ` Fe^(2+)` ion is :A. 5B. 6C. 3D. 4 |
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Answer» Correct Answer - B (b) `Fe^(2+) : 1 s^2, 2 s^2 2p^6, 3 s^2 3p^6 3 d^6`. |
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| 1124. |
The `(e)/(m)` value of proton isA. Less than `(e)/(m)` value of electronB. Equal to `(e)/(m)` value of electronC. Greater than `(e)/(m)` value of electronD. All the above |
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Answer» Correct Answer - A specific charge equal to charge to mass ratio |
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| 1125. |
The orbital angular momentum of an electron in `2s`-orbital isA. `(h)/(4pi)`B. zeroC. `(h)/(2pi)`D. `(sqrt(2)h)/(2pi)` |
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Answer» Correct Answer - B `L = sqrt(l(l+1)) (h)/(2pi)` |
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| 1126. |
An isotone of `._32^76Ge` isA. `._32^77Ge`B. `._33^77As`C. `._34^77Se`D. `._36^78Se` |
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Answer» Correct Answer - B::D `._33As^77` and `._34Se^78` have same number of neutrons =(A-Z) as `._32Ge^76` |
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| 1127. |
Azimuthal quantum number defines.A. `e//m` ratio of electronB. spin of electronC. angular momentum of electronD. magnetic momentum of electronic |
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Answer» Correct Answer - C ( c) Generally azimuthal quantum number defines angular momentum. |
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| 1128. |
Nitrogen laser produces a radiation at a wavelength of `337.1nm`. If the number of photons emitted is `5.6 xx 10^(24)`. Calculate the energy of this laserA. `3.33 xx 10^(6)J`B. `3.33 xx 10^(5)J`C. `1.56 xx 10^(6)J`D. `15.6 xx 10^(8)J` |
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Answer» Correct Answer - A `E = (Nhc)/(lambda)` |
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| 1129. |
An electron jumps from nth level to the first level .The correct face (s) about H atomic is//areA. Number of specteal lines ` = sum (n(n-1))/2`B. Number fo spectal lines ` = sum ( n-1)`C. Number fo spectal lines ` = sum ( n-1)`D. If ` n=4` the no.of spectral lines = six. |
| Answer» Correct Answer - A::B::D | |
| 1130. |
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 electronB. Alpha particles are positively chargedC. Alpha particles move with high velocityD. Most part of the atom is empty |
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Answer» Correct Answer - D Atom is mostly empty |
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| 1131. |
A particle `X` moving with a certain velocity has a debragile wave length of `1.72Å`, If particle `Y` has a mass of `50%` that of `X` and velocity `50%` that of `X`, debroglies wave length of `Y` will be `-`A. `3Å`B. `5.33Å`C. `6.88Å`D. `48Å` |
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Answer» Correct Answer - 3 `lambda=(h)/(mv)" "lambda_(x)=(h)/(m_(x)v_(x))" "lambda_(y)=(h)/(m_(y)v_(y))` `m_(y)=(m_(x))/(2)" "lambda_(y)=(V_(y))/(2)" "(lambda_(x))/(lambda_(y))=(1)/(4)` `lambda_(y)=1.72xx4=6.88A` |
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| 1132. |
Azimuthal quantum number defines.A. e/m ratio of electronB. Spin of electronC. Angular momentum of electronD. Magnetic momentum of electron |
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Answer» Correct Answer - C Generally azimuthal quantum number defines angular momentum |
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| 1133. |
The electronic configuration of copper `(._29Cu)` isA. `1s^2, 2s^2 2p^6 , 3s^2 3p^6 3d^9 , 4s^2`B. `1s^2, 2s^2 2p^6 , 3s^2 3p^6 3d^10 , 4s^1`C. `1s^2. 2s^2 2p^6 , 3s^2 3p^6, 4s^2 4p^6`D. `1s^2 , 2s^2 2p^6 , 3s^2 3p^6 3d^10` |
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Answer» Correct Answer - B 3d subshell filled with 10 electrons (fully-filled ) is more stable than that filled with 9 electrons. 1,4s electrons jumps into 3d subshell for more stability |
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| 1134. |
Photon of which light has maximum energy :A. RedB. BlueC. VioletD. Green |
| Answer» Correct Answer - C | |
| 1135. |
The charge on electron was discovered byA. J.J. ThomsonoB. R.A. MillikanC. RutherfordD. Chadwick |
| Answer» Correct Answer - B | |
| 1136. |
Given in hydrogen atom `r_(n),V_(n),E,K_(n)` stand for radius, potential energy, total energy and kinetic energy in `n^(th)` orbit. Find the value of U,v,x,y. `{:((A),u=(V_(n))/(K_(n)),(P),1,),((B),(1)/(r_(n))propE^(x),(Q),-2,),((C),r_(n)propZ^(y),(R),-1,),(,(Z="Atomic number"),,,),((D),v=("Orbital angular momentum of electron"),(S),0,),(,("in its lowest energy"),,,):}` |
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Answer» Correct Answer - A::B::C::D |
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| 1137. |
Many elements have non-integral atomic masses becauseA. They have isotopesB. Their isotopes have non-integral massesC. Their isotopes have different massesD. The constituents , neutrons , protons and electrons, combine to given fractional masses |
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Answer» Correct Answer - A::C Because they have isotopes with different masses. Their average atomic mass is the weighted mean of their presence in nature , e.g., `Cl^35` and `Cl^37` are present in ratio 3:1. So, `A=(35xx3+37xx1)/4 =35.5` |
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| 1138. |
Many elements have non-integral atomic masses becauseA. They have isotopesB. Their isoptopes have non-integral massesC. Their istopes have different massesD. The contituent-neutrons protons and electron-combine to give radaonal messes |
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Answer» Non integral atomic masses are due to isotopes which have difference masses Hence a and c are correct |
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| 1139. |
Many elements have non-integral atomic masses becauseA. they have isotopesB. their isotopes have non-intergral massesC. their isotopes have different massesD. the constituents, neutrons, protons and electrons combine to give rational masses |
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Answer» Non-integral atomic masses are due to isotopes which have different masses. So, (A) and (C ) are the correct answer. |
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| 1140. |
The atomic nucleus containsA. protonsB. neutronsC. electronsD. photons |
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Answer» Correct Answer - A::B |
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| 1141. |
A cylindrical source of light which emits radiation radially (from curved surface) only , placed at the centre of hollow, metallic cylindrical surface, as shown in diagram. The power of source is 90 watt and it emits light of wavelength `4000 "Å"` only . The emitted photons strike the metallic cylindrical surface which results in ejection of photoelectrons. All ejected photoelectrons reaches to anode (light source). The magnitude of photocurrent (in amp) is : `["Given": h=6.4xx10^(-34) J//sec]` |
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Answer» Correct Answer - 10 |
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| 1142. |
When an electron revolves in a stationary orbit thenA. it absorbs energyB. It gains kinetic energyC. it emits radiationD. Its energy remains constant |
| Answer» Correct Answer - D | |
| 1143. |
The value of planck constant isA. `6.62 xx 10^(-34)` Joule-secondB. `6.62 xx 10^(-27)` Joule-secondC. `6.62 xx 10^(-30)` Joule-secondD. `6.62 xx 10^(-24)` Joule-second |
| Answer» Correct Answer - B | |
| 1144. |
The emission spectrum of hydrogen is found to satisfy the expression for the energy change `Delta E` (in joules) such that `Delta E = 2.18 xx 10^(-18)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))J` where `n_(1) = 1,2,3,....` and `n_(2) = 2,3,4,...` The spectral lines correspond to Paschen series ifA. `n_1=1` and `n_2`=2,3,4B. `n_1=3` and `n_2`=4,5,6C. `n_1=1` and `n_2`=3,4,5D. `n_1=2` and `n_2`=3,4,5 |
| Answer» Correct Answer - B | |
| 1145. |
A triply ionized Be-atom has the same radius of `2^(nd)` orbitas that of ground state of H-atom. Th radius of an orbit is `r_n = (r_1 xx n^2)/(1)`.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. |
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Answer» Correct Answer - A (a) `r_n = (r_1 xx n^2)/(Z) fo r_(2_(Be^(3+))) = (r_1 xx 2^2)/(4)` and fo r `r_1 H = (r_1 xx 1^2)/(1)`. |
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| 1146. |
Calculate the wavelength of the first line in the balmer series of hydrogen spectrum |
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Answer» For the first in Balmer series `n_(1) = 2 and n_(2) = 3` so `bar v R_(H)[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))] cm^(-1)` `= 109677[(1)/((2)^(2)) - (1)/((3)^(2))]cm^(-1)` `= 109677 xx (5)/(36) cm^(-1)` `= 15232.9 cm^(-1)` `or lambda = (1)/(bar v) = (1)/(15.232.9)cm = 6.565 xx 10^(-5) cm = 656.5 nm` |
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| 1147. |
Which of the following has the maximum number of ampaired electrons ?A. `Mg^(2+)`B. `Ti^(2+)`C. `V^(3+)`D. `Ve^(2+)` |
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Answer» `Fe^(2+) = [Ar]3d^(6) rArr 4 unpairede` `Mg^(2+)` has zero uapaired `V^(2+) has 2` unpaired and `Ti^(3+)` has `1` unpaired |
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| 1148. |
Find the energy required to excite `1.10` L of hydrogen gas at `1.0` `Nm^(-2)` and `298`K to the first excited state of atomic hydrogen. The energy required for the dissociation of `H-H` bond is `436 kJ mol^(-1)`. Also calculate the minimum frequency of a photon to break this bond. |
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Answer» We know, Mole of ` H_2` present in one litre `= (PV)/(RT)` ` = ( 1 xx 1)/( 0. 0. 821 xx 298) = 0.04 09` Thus energy needed to break ` H-H ` bods in ` 0. 0 4 09` mole of ` H_2 = 0. 409 xx 436 = 17 . 83 k J` Also energy needed tor ecite one H-atom from 1 st to 2nd energy level ` = 13. 6 ( 1 - 1/4 ) eV = 10 . 2 eV` ` = 10 . 2 xx 1.6 xx 10^(-19) J` ` :.` Energy needed to ecite ` 0. 04 09 xx 2 xx 6.02 xx 10^(23)` atoms of h ` = 10 . 2 xx 1.6 xx 10^(-19) xx 0.0409 xx 2xx 6. 02 xx 10^(23) J` Thus , total energy needed ` = 17 . 83 + 80 . 36 = 98 . 19 k J` Energy required to break ( H -H ) bond ` = ( 436 xx 10^3)/( 6.03=23 xx 10^(23)) ` joule ` :. (( 436 xx 10^3 )/( 6.023 xx 10^(23)) = 6. 625 xx 10^(-34))v" " ( :. E = hv)` ` :. v = 10 . 93 xx 10^(14) sec ^(-1) or Hz`. |
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| 1149. |
Find the energy required to excite `1.10L` od hydrogen gas at `1.0` nm and `298 K` to the first excited state of atomic hydrogen .The energy required for the dissacitation of `H-H` bond is `436 kJ mol^(-1)`.Also calculate the minimum frequency of a photon to break this bond |
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Answer» Let as first the number of moles of hydrogen atom `n_(H_2) = (PV)/(KT) = (1 xx 1.10)/(0.0921 xx 298) = 0.045` Thus the energy required to break `0.045` and of `H_(2)(H-H) "bond")` is `0.045 xx 436 = 19.62 kJ` Now calculate the energy needed excite the H atom in the first excited sstate i.e. to `n = 2` `Delta E = 2.18 xx 10^(--18) xx Z^(2)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) J "atom"^(-1)` `Delta E = 2.18 xx 10^(-18) xx (1)^(2)((1)/(1^(2)) - (1)/(2^(2))) J "atom"^(-1)` `= 1.635 xx 10^(-18) J "atom"^(-1)` Number of H atom of `H_(2)` molecules)`xx 2 = (0.05 xx 6.02 xx 10^(23)) xx 2= 6.02 xx 10^(22)` The energy required to excite the given number of H atom is `6.02 xx 10^(22) xx 1.635 xx 10^(-18) J = 98.43 kJ` So, the total energy required is `19.62 + 98.43 = 118.05 kJ` Now energy required to break a single H-H bond is `(436 xx 10^(3))/(6.023 xx 10^(22)) = 7.238 xx 10^(-19)` = Energy supplied by the photon `rArr 7.238 xx 10^(-19) = hv = 6.026 xx 10^(-34)(v)` `:. v = (7.38 xx 10^(-19))/(6.626 xx 10^(-34))` `rArr v = 1.09 xx 10^(15) Hz` |
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| 1150. |
Calculate the energy (in KJ) required to excite one litre of hydrogen gas at 1 atm and 298 K to the first excited state of atomic hydrogen. The energy for the dissociation of `H-H` id 436 KJ `mol^(-1)`. Give your answer excluding decimal places. |
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Answer» Correct Answer - 97 |
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