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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.
| 101. |
Pertain to the following statement and figure The figure above shown level diagram of the hydrogen atom. Serveral transition are market as I, II, III, … The diagram is only indicative and not to scale. In which transition ia a Balmer series photon absorbed?A. IIB. IIIC. IVD. VI |
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Answer» Correct Answer - D For Balmer series , `n_(1) = 2, n_(2) = 3,4`,… (lower) (higher) Therefore, in transition (VI), proton of Balmer series is absorbed. |
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| 102. |
Pertain to the following statement and figure The figure above shown level diagram of the hydrogen atom. Serveral transition are market as I, II, III, … The diagram is only indicative and not to scale. The wavelength of the radiation involved in transition II is:A. `291 nm`B. `364 nm`C. `487 nm`D. `652 nm` |
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Answer» Correct Answer - C In transition II, `E_(2) = - 3.4 eV, E_(4) = - 0.58 eV` `Delta E = 2.55 eV` `Delta E = (hc)/(lambda) implies lambda = (hc)/(Delta E) = 487 nm` |
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| 103. |
A TV tube operates with a `20 kV` accelerating potential. What are the maximum-energy X-rays from the TV set? |
| Answer» The electrons in the TV tube have an energy of `20 keV`, and if these electrons are brought to rest by a collision in which one X-ray photon is emitted,the photon energy is `20 keV`. | |
| 104. |
Hydrogen `(H)`, deuterium `(D)`, singly ionized helium `(He^(+))` and doubly ionized lithium `(Li)` all have one electron around the nucleus. Consider `n = 2` to `n = 1` transition. The wavelength of emitted radiations are `lambda_(1), lambda_(2), lambda_(3)` and `lambda_(4)` respectively. then approximatelyA. `lambda_(1) = lambda_(2) = 4lambda_(3) = 9lambda_(4)`B. `4lambda_(1) = 2lambda_(2) = 2lambda_(3) = lambda_(4)`C. `lambda_(1) = 2lambda_(2) = 2sqrt(2)lambda_(3) = 3sqrt(2)lambda_(4)`D. `lambda_(1) = lambda_(2) = 2lambda_(3) = 3sqrt(2)lambda_(4)` |
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Answer» Correct Answer - A Using `DeltaE prop Z^(2) ( :. n_(1)` and `n_(2)` are same) `rArr (hc)/(lambda) prop Z^(2) rArr lambda Z^(2) =` constant `rArr lambda_(1) Z_(1)^(2) = lambda_(2) Z_(2)^(2) = lambda_(3) Z_(3)^(2) = lambda_(4) Z_(4)^(2)` `rArr lambda_(1) xx 1 = lambda_(2) xx 1^(2) = lambda_(3) xx 2^(2) = lambda_(3) xx 3^(3)` `rArr lambda_(1) = lambda_(2) 4 lambda_(3) = 9 lambda_(4)` |
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| 105. |
Hydrogen atom in ground state is excited by a monochromatic radiation of `lambda = 975 Å`. Number of spectral lines in the resulting spectrum emitted will beA. 3B. 2C. 6D. 10 |
| Answer» Correct Answer - C | |
| 106. |
The wavelength of the first line in blamer series in the hydrogen spectrum is `lambda`. What is the wavelength of the second line:A. `(20lambda)/27`B. `(3lambda)/16`C. `(5lambda)/36`D. `(3lambda)/4` |
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Answer» Correct Answer - A `1/lambda_(1)= R(1/4-1/9) Rightarrow lambda_(1)=(4xx9)/(5R)` Similarly `1/lambda_(2)=R(1/4-1/4^(2))` `Rightarrowlambda_(2)=16/(3R)=16/3xx(5lambda)/(4xx9)=20/(27lambda)` |
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| 107. |
The characteristic X-rays spectrum is emitted due to transition ofA. valence electrons of the atomB. inner electron of the atomC. nucleus of the atomD. both, the inner electrons and the nucleus of the atom |
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Answer» Correct Answer - B The characteristic x-rays are obtained due to the transition of electron from inner orbits. |
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| 108. |
50% of the X-ray coming from a Cooling tube is able to pass through a 0.1 mm thick aluminium foil. The potential difference between the target and the filament is increased. The thickness of aluminium foil, which will allow 50% of the X-ray to pass through, will beA. zeroB. `lt0.1mm`C. `0.1mm`D. `gt0.1mm` |
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Answer» Correct Answer - B With increase of potentiaal differece, X-ray of higher energy will be produced , to stop them , thicker foil is required. |
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| 109. |
A particle of mass `M` at rest dacays into two particle of masses `m_(1)` and `m_(2) `having non zero velocity . The ratio of the de Broglie wavelength . The ratio of the de Broglie wavelength of the particle `lambda , _(1) // lambda_(2)` isA. `(m_(1))/(m_(2))`B. `(m_(2))/(m_(1))`C. `1:1`D. `sqrt((m_(2))/(m_(1)))` |
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Answer» Correct Answer - C `lambda=(h)/(p)` Since the momenta of the two particles are equal `lambda` are same. |
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| 110. |
The de Broglie wavelength of an neutron corresponding to root mean square speed at `927^(@)C` is `lambda`. What will be the de Broglie wavelength of the neutron corresponding to root mean square speed at `27^(@)C`?A. `(lambda)/(2)`B. `lambda`C. `2 lambda`D. `4 lambda` |
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Answer» Correct Answer - C `K.E.` of neutron `E=(3)/(2)KT` `lambda_(d)=(h)/(p)=(h)/(sqrt(2mE))=(h)/(sqrt(2mxx(3)/(2)kT))` `rArr lambda_(2)=lambdasqrt(((927+273))/(27+273))= 2 lambda` |
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| 111. |
Let p and E denote the linear momentum and energy of a photon. If the wavelength is decreased,A. both `P` and `E` increasesB. `p` increases and `E` decreasesC. `p` decreases and `E` increasesD. both `p` and `E` decreases |
| Answer» Correct Answer - A | |
| 112. |
The follwing diagram indicates the energy levels of a certain atom when the system moves from `2E` level to `E`, a photon of wavelength `lambda` is emitted. The wavelength of photon produced during its transition from `(4E)/(3)` level to `E` is A. `3lambda`B. `3//4lambda`C. `lambda//4`D. `2lambda` |
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Answer» Correct Answer - A `(hc)/=2E-E=Eto1,(hc)/lambda_(2)=4/3E-E=E/3to2` `lambda_(2)/lambda_(1)=(E)/(E/3)=3/1` |
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| 113. |
The following diagram indicates the energy levels of a certain atom when the system moves from `4E` level to `E`. A photon of wavelength `lambda_(1)` is emitted. The wavelength of photon produced during its transition from `(7)/(3)E` level to `E` is `lambda_(2)`. the ratio `(lambda_(1))/(lambda_(2))` will be A. `(9)/(4)`B. `(4)/(9)`C. `(3)/(2)`D. `(7)/(3)` |
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Answer» Correct Answer - B |
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| 114. |
An alpha-particle accelerated through `V` volt is fired towards a nucleus . Its distance of closest approach is . `r` If a protons is accelerated through the same potential and fired towards the same nucleus , the distance of closest approach of proton will be .A. `r`B. `2r`C. `r/2`D. `r/4` |
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Answer» Correct Answer - A `K.E=PERightarrowq_(1)V=1/(4piepsilon_(0))(q_(1)q_(2))/r` independent of charge of the particle projected |
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| 115. |
The energy change is greatest for a hydrogen atom when its state changes fromA. `n = 2 to n = 1`B. `n = 3 to n = 2`C. `n = 4 to n = 3`D. `n = 5 to n = 4` |
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Answer» Correct Answer - A In this case, there is the widest energy gap. |
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| 116. |
photoelectron are emitted when `4000A^(0)` radiation is incident on a surface of work fuction `1.9eV`. these photoelectrons pass through a region has a- particles to form `He^(+)` ion, emitting a single photon in this process `He^(+)` ions thus formed are in their fourth excited state. Energy of the fourth excited state is approxA. `-4.2eV`B. `-2.2eV`C. `-3.2eV`D. `-1.2eV` |
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Answer» Correct Answer - B `K_(max)=(hc)/lambda-phi=1.18eV,E=-(13.6eV)` |
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| 117. |
Find the maximum magnitude of the linear momentum of a photoelectron emitted when light of wavelength 400 nm falls on a metal having work function 2.5e V. |
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Answer» Correct Answer - `(P^(2))/(2m)((1.24xx10^(4))/(4000)-2.5)eV=0.6eV` `P=sqrt(2xx9.1xx10^(-31)xx0.6xx1.6xx10^(-19))` `=4.2xx10^(-25)kg.m//s` |
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| 118. |
The wavelength of radiation emitted is `lambda_(0)` when an electron jumps from the third to the second orbit of hydrogen atom. For the electron jump from the fourth to the second orbit of hydrogen atom, the wavelength of radiation emitted will beA. `(16)/(25) lambda_(0)`B. `(20)/(27)lambda_(0)`C. `(27)/(20)lambda_(0)`D. `(25)/(16)lambda_(0)` |
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Answer» Correct Answer - B `(1)/(lambda) = R [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))] rArr (1)/(lambda_(3 rarr 2)) = R[(1)/((2)^(2)) - (1)/((3)^(2))] = (5R)/(36)` and `(1)/(lambda_(4 rarr 2)) = R [(1)/((2)^(2)) - (1)/((4)^(2))] = (3R)/(16)` `:. (lambda_(4 rarr 2))/(lambda_(3 rarr 2)) = (20)/(27) rArr lambda_(4 rarr 2) = (20)/(27) lambda_(0)` |
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| 119. |
if the electron in hydrogen orbit jumps form third orbit to second orbit, the wavelength of the emitted radiation is given byA. `lambda = ( R)/(6)`B. `lambda = (36)/(5 R)`C. `lambda = ( 6)/(R)`D. `lambda = (5 R)/(36)` |
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Answer» Correct Answer - B `(1)/(lambda) = R [(1)/(2^(2)) - (1)/(3^(2))] = (5 R)/(36) : lambda = (36)/(5 R)` |
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| 120. |
The rediation emitted when an electron jumps from `n = 3 to n = 2` orbit in a hydrogen atom falls on a metal to produce photoelectron. The electron from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of `(1//320) T` in a radius of `10^(-3) m`. Find (a) the kinetic energy of the electrons, (b) Work function of the metal , and (c) wavelength of radiation. |
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Answer» When charged particle moves perpendicular to a magnetic field , then magneticfield provides necessary centripetel force for circle path of radius`r` given by `(m v^(2))/( r) = q v B` `implies m v = q B r` As momentum, `p = mv = sqrt(2 m E_(K))`, where `E_(K)` is kinetic energy , therefore `sqrt(2 m E_(K)) = qBr` `:. E_(k) = ((qBr)^(2))/(2 m)` `{1.6 xx 10^(-19) xx ((1)/(320)) xx 10^(-3)}^(2)/(2 xx 9.1 xx 10^(-31) J` `= 1.374 xx 10^(-19)J = (1.374 xx 10^(-19))/(1.6 xx 10^(-19)) e V` `= 0.86 e V` Energy of photon released due to transition from `n = 3 to n = 2` in hydrogen atom. `epsilon = Delta E` `= Rhc ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = (13.6 e V) ((1)/(2^(2)) - (1)/(3^(2)))` `= 1.89 e V` Work function of metal `W = epsilon - E_(k) = 1.89 - 0.86 = 1.03 e V` c. Wavelength of emitted radition (photon) is given by `Delta E = (hc)/(lambda)` `implies lambda = (hc)/(Delta E) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(1.89 xx 1.6 xx 10^(-19))` `= 6.567 xx 10^(-7) m = 6567Å` |
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| 121. |
The ionization energy of a hydrogen like Bohr atom is `4` Rydberg. If the wavelength of radiation emitted when the electron jumps from the first excited state to the ground state is `N-m` and if the radius of the first orbit of this atom is `r-m` then the value of `(N)/(r )=Pxx10^(2)` then, value of `P`. (Bohr radius of hydrogen `=5xx10^(11)m,1` Rydberg `=2.2xx10^(16)J`) |
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Answer» Correct Answer - `12` (a) In energy units, `1` rydberg `=13.6 eV`. The energy needed to detach the electron is `4xx13.6 eV`. The energy in the ground state is, therefore, `E_(1) = -4xx13.6 eV`. The energy of the first excited state `(n=2)` is `E_(2)=(E_(1))/(4)=13.6 eV. DeltaE=40.8 eV`. The wavelength of the radiation emitted is `lambda=(hc)/(DeltaE)` (b) The energy of a hydrogen-like ion in ground state is `E=Z^(2)E_(0)` where `Z=` atomic number and `E_(0) = -13.6 eV`. Thus, `Z=2`. The radius of th first orbit is `(a_(0))/(Z)` where `a_(0)=5xx10^(-11)m`. Thus, `r=(a_(0))/(Z)=2.5xx10^(-11)m` |
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| 122. |
The wave number of energy emitted when electron jumps from fourth orbit to second orbit in hydrogen in `20,497cm^(-1)`. The wave number of energy for the same trasition in `He^(+)` isA. `5,099cm^(-1)`B. `20,497cm^(-1)`C. `40,994cm^(-1)`D. `81,988cm^(-1)` |
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Answer» Correct Answer - D `bar(v)=RZ^(2)(1/n_(1)^(2)(-1)/n_(2)^(2))_,bar(v)aZ^(2)` |
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| 123. |
A `He^(+)` ion is at rest and is in ground state. A necutron with intial kinetic energy`K` collides heads on with the `He^(+)` ion. Find minimum value of `K` so that there can be an inelastic collision between these two particals. |
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Answer» Here the loss during the collision can only be used to excite the atoms or electrons So , according to quantum mechaincs possible losses can be `{0, 40.8 e V, 48.36 e V}` (i) from `E_(n) = - 13.6 e V (Z^(2))/(n^(2))` Now according to Newtonion mechanics. Minimum loss `= 0`. Maximum loss will be for perfectly inelclastic collision. Let `v_(0)` be the inital speed of neutron and `v_(f)` be the final common speed So, by momentum conservation : `m v_(0) = m v_(f) + 4m v_(f) implies v_(f) = (v_(0))/(5)` where `m = "mass of meutron and mass of" He^(+)` ion =` 4 m`. Final kinetic energy of system: `KE = (1)/(2) m v_(f)^(2) + (1)/(2) 4 m v_(f)^(2)` `= (1)/(2) (5 m) = v_(0)^(2)/(25) = (1)/(5) ((1)/(2) m v_(0)^(2)) = (K)/(5)` maximum loss `= K - (K)/(5) = (4K)/(5)` So , loss will be `[0 (4K)/(5)]` For inelastic collition, there should be at least one common value other than zero in Eqs. (i) and (ii) `:. (4K)/(5) gt 40.8 e V implies K gt 51 e V` Hence maximum value of `K` for an inelastic collision , `K_(min) = 51 e V`. |
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| 124. |
An `H` atom in ground state is moving with initial kinetic energy `K` .It collides head on with `He ^(+)` ion in ground state kept at rest but free to move. Find minimum value of `K` so that both the particles can excite to their first excited state.A. `63.75eV`B. `51eV`C. `54.4eV`D. `13.05eV` |
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Answer» Correct Answer - A Max . Loss In `KE` of the system take place in perfectly inelastic collision . From conservation of linear momentum ` "mu"+0=(m+M)VRightarrow=("mu")/(m+M) ` Max. loss in `KE` ` DeltaK=1/2"mu"^(2)-1/2(M+m)V^(2)=1/2"mu"^(2)(M/(M+m))` For both to get excited to their excited states, energy required is `DeltaE=10.2+10.2xx4=51eV`. therefore For both to get excited ` DeltaKgeDeltaERightarrow1/2"mu"^(2)(M/(M+m))geDeltaE` `Rightarrow 1/2"mu"geDeltaE(M+m)/m eVRightarrowKE_(min)=DeltaE(1+m/M)` ` i.e KE_(min)=51(1+1/4)=63.75eV` |
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| 125. |
A neutron moving with speed a makes a heat on collision with a hydrogen atom in ground state kept at rest .Find the minimum kinetic energy of the neutron for which inelastic (completely or partially) collision may take plane at the mass of neutron = mass of hydrogen `= 1.67 xx 10^(-27) kg` |
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Answer» Suppose the neutron and hydrogen atom move at speed `v_(1)` and `v_(2)` after the collision. The collision on will be inelastic if a part of the kinetic energy is used to excite the atom. Suppose an energy `DeltaE` is used in this way. Using conservation of linear momentum and energy. `mv=mv_(1)+mv_(2)`.....(i) and `(1)/(2)mv^(2)=(1)/(2)mv_(1)^(2)+(1)/(2)mv_(2)^(2)+DeltaE`......(ii) From (i), `v^(2)=v_(1)^(2)+v_(2)^(2)+2v_(1)v_(2)`, From (ii) `v^(2)=v_(1)^(2)+v_(2)^(2)+(2DeltaE)/(m)` thus,`2v_(1)v_(2)=(2DeltaE)/(m)` Here, `(v_(1)-v_(2))^(2)-4v_(1)v_(2)=v^(2)-(4DeltaE)/(m)` As `v_(1)-v_(2)` must be real, `v^(2)-(4DeltaE)/(m) gt 0` or `(1)/(2)mv^(2) gt 2DeltaE` The minimum energy that can be absorbed by the hydrogen atom in ground state to go in an excited state is `10.2 eV`. Thus, the minimum kinetic energy of the nerutron needed for an inelastic collision is `(1)/(2)mv_(min)^(2)=2xx10.2eV=20.4 eV` |
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| 126. |
A neutron moving with a speed `v` makes a head-on collision with a hydrogen in ground state kept at rest which inelastic collision will be take place is (assume that mass of photon is nearly equal to the mass of neutron)A. `10.2eV`B. `20.4eV`C. `12.1eV`D. `16.8eV` |
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Answer» Correct Answer - B Let V=speed of neutron before collision `v_(1)`=speed of neutron after collision ` v_(2)`= speed of proton or hydrogen atom after collision and `DeltaE`= energy of exitation. From conservation of linear momentum `mv=mv_(1)+mv_(2)` ...(1) From conservation of energy ` 1/2mv^(2)=1/2mv_(1)^(2)+1/2mv_(2)^(2)+DeltaE` ...(2) from `1`and `2`As `v_(1)-v_(2)` must be real `v^(2)-4(DeltaE)/mge0` ` Rightarrow1/2mv^(2)=2DeltaE=2xx10.2=20.4eV` |
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| 127. |
A neutron moving with a speed `v` makes a head-on collision with a hydrogen in ground state kept at rest which inelastic collision will be take place is (assume that mass of photon is nearly equal to the mass of neutron)A. `10.2 eV`B. `20.4 eV`C. `12.1 eV`D. `16.8 eV` |
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Answer» Correct Answer - B Let `v = ` speed of neutron before collision, `v_(1) = ` speed of neutron after collision, `v_(2) = ` speed of photon or hydrogen atom after collision, and `Delta E=` energy of excitation From conservartion of linear momentum, `mv = mv_(1) + mv_(2)` (i) From conservation of energy `(1)/(2) mv^(3) = (1)/(2) mv_(1)^(2) + (1)/(2) mv_(2)^(2) + Delta E` (ii) From eq. (i), `v^(2) = v_(1)^(2) + v_(2)^(2) + 2 v_(1) v_(2)` From eq. (ii), `v^(2) = v_(1)^(2) + v_(2)^(2) + (2 Delta E)/(m)` `:. 2 v_(1) v_(2) = (2 Delta E)/(m)` `:. (v_(1) - v_(2))^(2) = (v_(1) + v_(2))^(2) - 4v_(1) v_(2)` `implies (v_(1) - v_(2))^(2) = V^(2) - 4 (Delta E)/(m)` As `v_(1) - v_(2)` must be real, therefore `v^(2) - 4 (Delta E)/(m) ge 0` or `(1)/(2) mv^(2) ge 2 Delta E` The minimum energy that can be obsorbed by hydrogen atom in the ground state go to into excited is `10.2 eV`. Therefore, `(1)/(2) mv_(min)^(2) = 2 xx 10.2 eV` `= 20.4 e V` |
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| 128. |
In an excited state of hydrogen like atom an electron has total energy of `-3.4 eV`. If the kinetic energy of the electron is E and its de-Broglie wavelength is `lambda`, thenA. `E = 6.8 eV: lambda = 6.6 xx 10^(-10) m`B. `E = 3.4 eV: lambda = 6.6 xx 10^(-10) m`C. `E = 3.4 eV: lambda = 6.6 xx 10^(-11) m`D. `E = 6.8 eV: lambda = 6.6 xx 10^(-11) m` |
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Answer» Correct Answer - B Potential energy `= - 2 xx` kinetic energy = `- 2 E` Total energy `= - 2 E + E = - 3.4 eV = - E` `E = 3.4 eV` `P = "momentum"`, m = mass of electron or `P = sqrt(2 m E)` `= sqrt(2 xx 9.1 xx 10^(-31) xx 3.4 xx 1.6 xx 10^(-19)) = 10^(-24)` de Brogile wavelength `lambda = (h)/(P) = (h)/(sqrt(2 m E)) = 6.6 xx 10^(-10) m` |
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| 129. |
An electtron and a photon have same wavelength . If `p` is the moment of electron and `E` the energy of photons, the magnitude of `p//E in S I` unit isA. `3.0 xx 10^(8)`B. `3.33 xx 10^(-9)`C. `9.1 xx 10^(-31)`D. `6.64 xx 10^(-34)` |
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Answer» Correct Answer - B `lambda = (h)/(p)` (for electron) or `p = n//h` and `E = (hc)/(lambda)` (for photon) `:. (p)/(E) = (1)/( c) = (1)/(3 xx 10^(8)) = 3.33 xx 10^(-9) s m^(-1)` |
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| 130. |
In hydrogen and hydrogen-like atom , the ratio of `E_(4 n) - E_(2 n) and E_(2 n) - E_(n)` varies with atomic nimber `z` and principal quantum number`n` asA. `(z^(2))/(n^(2))`B. `(z^(4))/(n^(4))`C. `(z)/(n)`D. none of these |
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Answer» Correct Answer - D `(E_(4 n) - E_(2 n))/ (E_(2 n) - E_(n)) = ((E_(1))/(16 n^(2)) - (E_(1))/(4 n^(2)))/((E_(1))/(4 n^(2)) - (E_(1))/(n^(2))) = (1)/(4) = constant` |
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| 131. |
A charge of magnitude `3e` and mass `2m` is moving electric field `vecE`. The acceleration imparted to the charge isA. `2Ee//3m`B. `3Ee//2m`C. `2m//3Ee`D. `3m//2Ee` |
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Answer» Correct Answer - B Acceleration `a = (QE)/(m) = ((3e)E)/(2m)` |
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| 132. |
An electron initially at rest, is accelerated through a potential difference of `200` volt, so that it acquires a velocity `8.4 xx 10^(6)m//s`. The value of `e//m` of elctronA. `2.76 xx 10^(12) C//kg`B. ` 1.76 xx 10^(11)C//kg`C. `0.76 xx 10^(12)C//kg`D. None of these |
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Answer» Correct Answer - B `(1)/(2)mv^(2) = eV rArr (e)/(m) = (v^(2))/(2V) = ((8.4 xx 10^(6))^(2))/(2 xx 200)` `= 1.76 xx 10^(11) (C )/(kg)` |
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| 133. |
In the above question, if the scientist continues taking data at higher photon energies he will find the next major 'dip' in the intensity graph at what photon energy ?A. `(1)/(9) E_(0)`B. `(8)/(9) E_(0)`C. `3 E_(0)`D. `9 E_(0)` |
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Answer» Correct Answer - B `E_(photon) = E_(3) - E_(1) = - ( E_(0))/( 3^(2)) - ((-E_(0))/(1^(2))) = (8)/(9) E_(0)` |
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| 134. |
In the Angular process as atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (this is called an Augar elecrons ) . Assuming the nuclesu to be massive , calculate the kinetic energy of an `n=4` Augar electron emitted by chromium by absorbing the energy from a `n=2` to `n=1` transition .A. `5.4keV`B. `10.2KeV`C. `13.6KeV`D. `7.3KeV` |
| Answer» Correct Answer - A | |
| 135. |
A gas of hydrogen - like atoms can absorb radiations of `698 eV`. Consequently , the atoms emit radiation of only three different wavelengths . All the wavelengthsare equal to or smaller than that of the absorbed photon. a Determine the initial state of the gas atoms. b Identify the gas atoms c Find the minimum wavelength of the emitted radiation , d Find the ionization energy and the respective wavelength for the gas atoms. |
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Answer» (a) `(n(n-1))/(2)=3` `:. n=3` i.e., after excitation atom jumps to second excited state. hence `n_(f)=3`. So `n_(1)` can be `1` or `2` If `n_(1)=1` then energy emitted is either equal to, greater than or less than the energy absorbed. Hence the emitted wavelength is either equal to, less than or greater than the absorbed wavelength. Hence `n!=2`. If `n_(1)=2`, then `E_(e )geE_(a)` Hence `lambda_(e ) lelambda_(0)` (b) `E_(3)-E_(2)=68ev` `:. (13.6)(Z^(2))((1)/(4)-(1)/(9))=68` `:. Z=6` (c ) `lambda_(min)=(12400)/(E_(3)-E_(1))=(12400)/((13.6)(6)^(2)(1-(1)/(9)))=(12400)/(435.2)=28.49` (d) Ionization energy `= (13.6)(6)^(2)=489.6eV` `lambda=(12400)/(489.6)=25.33A` |
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| 136. |
The first excitation potential of a hypothetical hydrogen- like atom is `15 V`. Find the third excitation potential of the atom.A. `V` electron voltB. `(3 V)/(4)` electron voltC. `(4 V)/(3)` electron voltD. cannot be calculate by given information |
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Answer» Correct Answer - C first excitation energy is `R h c ((1)/(1^(2)) - (1)/(2^(2))) = R h c (3)/(4)` ` (3)/(4) R h c = V eV` `:. R h c = (4 V)/(3) eV` |
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| 137. |
A proton and electron, both at rest initially, combine to form a hydrogen atom in ground state. A single photon is emitted in this process. What is the wavelength? |
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Answer» Correct Answer - `912 Å` `lambda=(12400eV)/(13.6)= 911.76Å~~912Å` |
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| 138. |
The first excitation potential of a hypothetical hydrogen- like atom is `15 V`. Find the third excitation potential of the atom. |
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Answer» Let energy of ground state `=E_(0)` `E_(0)= -13.6Z^(2)eV` and `E_(n)=(E_(0))/(n^(2))` `n=2, E_(2)=(E_(0))/(4)` given `(E_(0))/(4)-E_(0)=15` `-(3E_(0))/(4)=15` for `n=4, E_(4)=(E_(0))/(16)` third exicitation energy `=(E_(0))/(16)-E_(0)` `=-(15)/(16)E_(0)=-(15)/(16).((-4xx15)/(3))` `=(75)/(4)eV` `:.` third exclination potential is `(75)/(4)V` |
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| 139. |
When a particle is restricted to move along x-axis between `x=0` and `x=a`, where `alpha` if of nenometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends `x=0` and `x=a`. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as `E=(p^2)/(2m)`. Thus the energy of the particle can be denoted by a quantum number `n` taking values 1,2,3, ...(`n=1`, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving along the line from `x=0` to `x=alpha`. Take `h=6.6xx10^(-34)Js` and `e=1.6xx10^(-19)` C. Q. If the mass of the particle is `m=1.0xx10^(-30)`kg and `alpha=6.6nm`, the energy of the particle in its ground state is closest toA. `0.8nmV`B. `8meV`C. `80meV`D. `800meV` |
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Answer» Correct Answer - B `E=h^(2)/(2m(4a^(2))) ("in ground state" lambda/2=a)approx8meV` |
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| 140. |
When a particle is restricted to move along x-axis between `x=0` and `x=a`, where `alpha` if of nenometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends `x=0` and `x=a`. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as `E=(p^2)/(2m)`. Thus the energy of the particle can be denoted by a quantum number `n` taking values 1,2,3, ...(`n=1`, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving along the line from `x=0` to `x=alpha`. Take `h=6.6xx10^(-34)Js` and `e=1.6xx10^(-19)` C. Q. If the mass of the particle is `m=1.0xx10^(-30)`kg and `alpha=6.6nm`, the energy of the particle in its ground state is closest toA. `0.8 me V`B. `8 meV`C. `80 meV`D. `800 meV` |
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Answer» Correct Answer - B `E = (n^(2)h^(2))/(8 ma^(2))` For ground state `n = 1 rArr E_(1) = ((1)^(2)(6.6 xx 10^(-34))^(2))/(8 xx 10^(-30) xx (6.6 xx 10^(-9))^(2) xx 1.6 xx 10^(-19)) = 8 m eV` |
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| 141. |
Positronium is just like a H-atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium ?A. `-136eV`B. `-6.8eV`C. `-3.4eV`D. `-1.51eV` |
| Answer» Correct Answer - B | |
| 142. |
A hydrogen atom in a state of binding energy` 0.55eV` make a trasisition to a state of excitation energy of `10.2 eV` (i) what is the initial state of hydrogen atom? (ii) what is the final state of hydrogen atom? (iii) what is the wavelength of the photon emitted? |
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Answer» (i) Let `n_(1)` be initial state of electron. Then ` E_(1)=-(13.6)/(n_(1)^(2))eV`, Here `E_(1) =- 0.85 eV`, therefore` -0.85=-(13.6)/(n_(1)^(2))` or `n_(1)=4` (ii) Let `n_(2)` be the final excition state of the electron. Since excition energy is a always measured with respect to the ground state, therefore ` DeltaE =13.6[1-1/n_(2)^(2) ]` here `DeltaE = 10.2 eV`, therefore, `10.2 = 13.6 [1-(1)/(n_(2)^(2))]` or `n_(2)=2` Thus , the electron jump from `n_(1) =4 to n_(2) =2` . (iii) the wavelength of the photon emitted for a transition between `n_(1) =4 n_(2) =2`, given by ` 1/lambda=R_(oo)[1/n_(2)^(2)-1/n_(1)^2] (or) 1/lambda=1.09xx10^(7)[1/2^(2)-1/4^(2)]= `4860 Å` |
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| 143. |
STATEMENT - 1 If the accelerating potential in an X - rays tube is increased, the wavelength of the characteristic X- rays do not change .STATEMENT -2 When an electron beam strikes the target in an X- rays tube, part of the kinectic energy is converted into X - rays energy . |
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Answer» Correct Answer - B Characteristic `X`-ray depend only upon the torget metar. |
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| 144. |
Two H atoms in the ground state collide in elastically. The maximum amount by which their combined kinetic energy is reduced isA. `10.20eV`B. `20.40eV`C. `13.6eV`D. `27.2eV` |
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Answer» Correct Answer - A The total energy associate with the two H-atoms in the ground state collide in elastically `=((13.6)/2^(2))+(13.6)=17.0eV` "therefore" , maximum loss of their combined kinetic energy `=27.2-17.0=10.2eV` |
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| 145. |
Which of the following is in the increasing order for penetrating powerA. `gamma`-raysB. `beta`-raysC. `alpha`-raysD. cathode rays |
| Answer» Correct Answer - A | |
| 146. |
In a hydrogen-like atom , an electron is orbating in an orbit having quantum number `n`. Its frequency of revolution is found to be `13.2 xx 10^(15)` Hz. Energy required to move this electron from the atom to the above orbit is `54.4 eV`. In a time of ? nano second the electron jumps back to orbit having quantum number`n//2`.If `tau` be the average torque acted on the electron during the above process, then find `tau xx 10^(27)` in Nm. (given: `h//lambda = 2.1 xx 10^(-34) J-s`, friquency of revolution of electron in the ground state of `H` atom `v_(0) = 6.6 xx 10^(15)` and ionization energy of `H` atom `(E_(0) = 13.6 eV)` |
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Answer» Correct Answer - 15 `v = v_(0) (Z^(2))/(n^(3)) implies (Z^(2))/(n^(3)) = 2 (i)` `E = E_(0) (Z^(2))/(n^(2)) implies (Z^(2))/(n^(3)) = 4 (ii)` Solving (i) and (ii), `n = 2, Z = 4` `L = mvr = (nh)/(2 pi), Delta L = tau Delta r = Delta n (h)/(2 pi)` `tau = (Delta n)/(Delta t) xx (h)/(2 pi) = (1)/(7 xx 10^(-9)) xx (1)/(2) xx 2.1 xx 10^(-34) = (2.1)/(14) xx 10^(-25)` `tau = 10^(27) = (2.1)/(14) xx 10^(-25) xx 10^(27) = 15` |
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| 147. |
In the Bohr model of the hydrogen atom, the ratio of the kinetic energy to the total energy of the electron in a quantum state `n` is …….. |
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Answer» Correct Answer - d `KE = (KZe^(2))/(2 r)` and Total energy , `TE = (-KZe^(2))/(2 r) implies (|KE|)/(|TE|) = 1` |
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| 148. |
The velocity of an electron in the first orbit of `H` atom is `v` . The velocity of an electron in the second orbit of `He^(+)` isA. `2 v`B. vC. `(v)/(2)`D. `(v)/(4)` |
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Answer» Correct Answer - B `v_(n) = alpha ((eZ)/(n))` , were `alpha = (e^(2))/(2 h epsilon_(0)c)` is the constant `(alpha = (1)/(137))` `v_(He) = alpha (c(2))/(2) = alpha c` and `v_(H) = alpha (c(1))/(1) = alpha c = v_(He^+)` |
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| 149. |
The recoil speed of a hydrogen atom after it emits a photon is going form n=5 state to n =1 state is ….. m/s.A. `4.718 m s^(-1)`B. `7.418 m s^(-1)`C. `4.178 m s^(-1)`D. `7.148 m s^(-1)` |
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Answer» Correct Answer - C `E = R_(oo) hc (1 - (1)/(25))` Momentum of the photon emitted is, `p` `m v = p` `(p)/(m) = ((1.097 xx 10^(7)) (6.626 xx 10^(-34)) 24)/((25) (1.67 xx 10^(-27))) = 4.178 m s^(-1)` |
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| 150. |
The recoil speed of a hydrogen atom after it emits a photon is going form n=5 state to n =1 state is ….. m/s. |
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Answer» Correct Answer - 4 `6 (1 - (1)/(25)) eV = 13.0 eV` `E//c = mv ("inomomentum conserved")` `v = (E)/(mc) = E_(photon) = (13.13 (1.6 xx 10^(-19)))/((1.67) (10^(-27)) (3) (10^(8))) = 4 m//s` |
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