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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.
| 151. |
If an electron jumps from `1`st orbital to `3`rd orbital, than it will.A. absorb energyB. release energyC. no gain of energyD. none of these |
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Answer» Correct Answer - A When an electron jumps from the orbit of lower energy (`n = 1)` to the orbit of higher energy `(n = 3)`, energy is absorbed. |
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| 152. |
A gas of identical hydrogen-like atoms has some atoms in the lowest in lower (ground) energy level `A` and some atoms in a partical upper (excited) energy level `B` and there are no atoms in any other energy level.The atoms of the gas make transition to higher energy level by absorbing monochromatic light of photon energy `2.7 e V`. Subsequenty , the atom emit radiation of only six different photon energies. Some of the emitted photons have energy `2.7 e V` some have energy more , and some have less than `2.7 e V`. a Find the principal quantum number of the intially excited level `B` b Find the ionization energy for the gas atoms. c Find the maximum and the minimum energies of the emitted photons. |
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Answer» a. Since only six different transition take place , the final state is `n = 4`. The energy level of hydrogen atom are given by `E_(n) = - (13.6)/(4^(2)) e V` If `n_(B)` is the principal quantum number of the initially excited state `B` , than `E_(4) - E_(n_B) = - (13.6)/(4^(2)) - ((- 13.6)/(n_(B)^(2)))` ` = 13.6 [(1)/(n_(B)^(2)) - (1)/(16)]` `E_(4) - E_(n_B) = 2.7 e V` `2.7 = 13.6 [(1)/(n_(B)^(2)) - (1)/(16)]` which gives `n_(B) = 2`. ("Rounding off to nearst integet")` b. The transition energy is numerically equal to the ground state energy `E_(1)` of level A. `E_(4) = (E_(1))/(16), E_(2) = (E_(1))/(4)` `E_(4) - (E_(2) = E_(1))/(16) - (E_(1))/(4)` `2.7 e V = - (3)/(16) E_(1)` `E_(1) = - 14.4 e V` Thus , the ionization energy of the given atom is `14.4 e V` c. Maximum energy of the emitted photon is for the electron transition `n = 4` to `n = 1`, i.e., `E_(4) - E_(3) = (E_(1))/(16) - E_(1) = - (15)/(16) E_(1)` `= (15)/(16) xx (14.4) = 13.5 e V` Thus, the maximum energy of the emitted photon is `13.5 e V`. Maximum energy of the emitted photon corresponds to the transition `n = 4 to n = 3`, i.e., `E_(4) - E_(3) = (E_(1))/(16) - E_(1)/(9) = - (7)/(144) E_(1)` `= - (7)/(144) xx (-14.4) = 0.7 e V` |
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| 153. |
In the lowest energy level of hydrogen atom, the electron has the angular momentumA. `pi/h`B. `h/pi`C. `h/2pi`D. `(2pi)/h` |
| Answer» Correct Answer - C | |
| 154. |
In the lowest energy level of hydrogen atom, the electron has the angular momentumA. `pi//h`B. `h//pi`C. `h//2pi`D. `2pi//h` |
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Answer» Correct Answer - C `mvr = (nh)/(2pi)`, for `n = 1` it is `(h)/(2pi)` |
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| 155. |
The wavelength of `L_alpha` line in X-ray spectrum of `._76Pt` is 1.32Å. The wavelength of `L_alpha` line in X-ray spectrum of another unknown element is 4.17Å. If screening constant for `L_alpha` line is 7.4, then atomic number of the unknown element isA. 78B. 47C. 40D. 35 |
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Answer» Correct Answer - B `sqrt(v) =a(Z-b)=sqrt(C/(lambda)) =a(Z-b)rArr sqrt((lambda_(2))/(lambda_(1)))=(a(78-7.4))/(a(Z_(2)-7.4))` `Z_(2)=47` |
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| 156. |
The minimum kinetic energy required for ionization of a hydrogen atom is `E_(1)` in case electron is collided with hydrogen atom , it is `E_(2)` if the the hydrogen ion is collided and `E_(1)` when helium ion collided . Then.A. `E_(1) = E_(2) = E_(3) `B. `E_(1) gt E_(2) gt E_(3) `C. `E_(1) lt E_(2) lt E_(3) `D. `E_(1) gt E_(3) gt E_(2) ` |
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Answer» Correct Answer - C Assuming that ionization occurs as a result of a completely inelastic collision, we can write `m v - 0 = (m + m_(H)) u` where `m` is that mass of incident partical , `m_(H)` the mass of hydrogen atom , `v_(0)` the initial velocity of incident particals , and `u` the final common velocity of the partical after collision. Prior to collision, the `KE` of the incident partical was `E_(0) = (m v_(0)^(2))/(2)` The total kinetic energy after collision `E = ((m + m_(H)) u^(2))/(2) = (m^(2)v_(0)^(2))/(2 (m + m_(H)))` The decrease kinetic energy must be eqaul to ionization energy. Therefore `E_(1) = E_(0) - E = ((m_(H))/(m + m_(H))) E_(0)` i.e., `(E_(1))/(E_(0)) = (1)/(1 + (m)/(m_(H))` i.e., the greater the mass `m`, the smaller the fraction of initial kinetic energy that be used for ionization. |
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| 157. |
An X-ray tube is operated at `50 kV and 20 m A`. The target material of the tube has mass of `1 kg` and specific heat `495 J kg ^(-1) `^(@)C^(-1)`. One perent of applied electric power is converted into X-rays and the remaining energy goes into heating the target. Then,A. a suitable target metrial must have hight melting temprature.B. a suitable target metrial must have low thermal conductivity.C. the average rate of rice of temprature of the target would be `2 ^(@) C s^(-1)`D. the minimum wavelength of X-rays emitted is about `0.25 xx 10^(-10) m` |
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Answer» Correct Answer - A::C Power loss increases the temprature. |
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| 158. |
If the operating potential in an X-ray tube is increased by 1%, by what percentage does the cutoff wavelength decrease? |
| Answer» Correct Answer - approximately `0.1%` | |
| 159. |
A beam of electron accelerated by a large potential difference `V` is mode of strike a metal target of produce X-ray . For which of the following value of `V` , will the resulting X-ray have the lower minimum wavelength? |
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Answer» Correct Answer - D `lambda_(min) = (hc)/(eV_(max))` |
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| 160. |
An X-ray tube operating at 50kV converts `1%` energy in the form of X-ray. If the amount of heat produced is 495 watt, then the number of electrons colliding with the target per second isA. `6.25xx10^(16)`B. `4.15 xx10^(16)`C. `3.2xx10^(16)`D. `1.2xx10^(16)` |
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Answer» Correct Answer - A `N=495/(50 KeV) =6.25 xx10^(16)` |
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| 161. |
A beam of electron accelerated by a large potential difference `V` is mode of strike a metal target of produce X-ray . For which of the following value of `V` , will the resulting X-ray have the lower minimum wavelength?A. 10 kVB. 20 kVC. 30 kVD. 40 kV |
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Answer» Correct Answer - D `lambda` will be minimum for maximum acceleration voltage |
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| 162. |
The ground state energy of hydrogen atom is-13.6 eV. What are the kinetic and potential energies of the electron in this state ? |
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Answer» Total energy `=(1)/(2)mv^(2)+((-Kze^(2))/(r )) " "( because (mv^(2))/(r )=(Kze^(2))/(r^(2)))` `=(Kze^(2))/(2 r)` `= -13.6 eV` (for `n=1, z=1`) Kinetic energy `=(1)/(2) mv^(2)=(Kze^(2))/(2r)=13.6eV` Potential energy `=(-Kze^(2))/(r )=2xx13.6= -27.2 eV` |
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| 163. |
A monochromatic light source of frequency `6xx10^(14)Hz` is emitting energy at the rate `2xx10^(-3)J//s`. Calculate the number of photons emitted per seconds by source. |
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Answer» Given that `v=6xx10^(14)Hz` `P=Q=2xx10^(-3)J//s` number of photons emitted per sec by the source. `rArr n(P)/(nv)=(2xx10^(-3))/(6.62xx10^(-34)xx6xx10^(14))` `=0.5035xx10^(-15)` per second. |
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| 164. |
The ratio of the speed of the electron in the first Bohr orbit of hydrogen and the speed of light is equal to (where `e, h and c` have their usual meaning in cgs system)A. `2 pi h // e^(2)`B. `e r^(2) h// 2 pi c`C. `e ^(2) c// 2 pi h`D. `2 pi e^(2)// h c` |
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Answer» Correct Answer - D `v_(n) = k (2 pi e^(2))/(n h)` We know that in cgs system`k = 1` `:. v_(n) = (2 pi e^(2))/(n h) implies v_(1) = (2 pi e^(2))/(h)` So `(v_(1))/(c) = k (2 pi e^(2))/(c h)` |
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| 165. |
Suppose two deuterons must get as close as `10^(-14) m` in order for the nuclear force to overcome the repulsive electrostatic force. The height of the electrostatic harrier is nearest toA. `0.14 M e V`B. `2.3 M e V`C. `1.8 xx 10 M e V`D. `0.56 M e V` |
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Answer» Correct Answer - A Barrier height `= (1)/(4 pi epsilon_(0)) (e^(2))/(r_(e)) J = (1)/(4 pi epsilon_(0)) (e)/(r_(e)) eV` `= (9 xx 10^(9) xx 1.6 xx 10^(-19))/(10^(-14)) eV = 1.44 xx 10^(5) eV` |
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| 166. |
In hydrogen atom, electron makes transition from `n = 4` to `n = 1` level. Recoil momentum of the `H` atom will beA. `3.4 xx 10^(-27)N-sec`B. `6.8 xx 10^(-27)N-sec`C. `3.4 xx 10^(-24)N-sec`D. `6.8 xx 10^(-24)N-sec` |
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Answer» Correct Answer - B Recoil momentum `=` momentum of photon `= (h)/(lambda)` `= hR ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = (hR xx 15)/(16) - 6.8 xx 10^(-27) N xx sec` |
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| 167. |
An electron in `H` atom makes a transition from `n = 3 to n = 1`. The recoil momentum of the `H` atom will beA. `6.45 xx 10^(-27) N s`B. `6.8 xx 10^(-27) N s`C. `6.45 xx 10^(-24) N s`D. `6.8 xx 10^(-24) N s` |
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Answer» Correct Answer - A The recoil momentum of atom is same as that of photon but in opposite direction. hence recoil momentum: `P = (E)/( c) = (12.09 xx 1.6 xx 10^(-19))/(3 xx 10^(8) N s = 6.45 xx 10^(-27) N s` note that almost whole of the energy will be carried away by the photon because it is very light in comparoson to `H` atom. |
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| 168. |
In a mixture of `H- He^(+)` gas (`He+ `is singly ionized He atom), `H` atom and `He+` ions are excited to their respective first excited state. Subsequently `H` atoms transfer their total excitation energy to `He+` ions (by collisions) Assume that the bohr model of atom is exactly valid. The wavelength of light emitted in the visible region by `He+` lons after collisions with `H` atoms is -A. `6.5xx10^(-7)m`B. `5.6xx10^(-7)m`C. `4.8xx10^(-7)m`D. `4.0xx10^(-7m` |
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Answer» Correct Answer - C The wavelength corresponding to transition from `n=4` to `n=3` in `He^(+)` correspond to visible region. |
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| 169. |
In a mixture of `H- He^(+)` gas (`He+ `is singly ionized He atom), `H` atom and `He+` ions are excited to their respective first excited state. Subsequently `H` atoms transfer their total excitation energy to `He+` ions (by collisions) Assume that the bohr model of atom is exactly valid. The ratio of the kinetic energy of the `n = 2` electron for the `H` atom to the of `He^(+)` lon is -A. `1/4`B. `1/2`C. `1`D. `2` |
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Answer» Correct Answer - A `K.E =-T.E=13.6Z^(2)/n^(2)eV` |
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| 170. |
An electron collides with a hydrogen atom in its ground state and excites it to `n = 3` ,. The energy gives to hydrogen aton n this inclastic collision is [Neglect the recoiling of hydrogen atom]A. `10.2 eV`B. `12.1 eV`C. `12.5 eV`D. none of these |
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Answer» Correct Answer - B The energy taken by hydrogen atom corresponds to its transition from n = 1 to n = 2 state. Delta E , (given to hydrogen atom) `= 13.6 ((1 - (1)/(9)) = 13.6 xx (8)/(9) = 12.1 eV` |
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| 171. |
Mark out the correct statement regarding X-rays.A. When fast moving electrons strike the metal target, they enter the metal larged and in a very short time span come to rest , and thus an electromagnetic charged electron produces electromagnetic waves (X-ray).B. Characteristic X-rays are produced due to transition of an electron from higher energy levels to vacant lower energy levels.C. X-ray spectrum is a discrete spectra just like hydrogen spectra.D. Both (a) and (b) are correct. |
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Answer» Correct Answer - B Option (a) explains the production of X-ray on the basis of electromagnetic theory of light , which is not above to explain the characteristic X-ray and cut-off wavelength , so option (a) is wrong. Option (b) correctly explain the production of characteri-stic X-ray. Option ( c) is wrong as X-ray spectra is a continuous spectra having some peak representing characterstic x-rays. |
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| 172. |
Define the following in photoelectric effect phenomenon (a) (i) work function (ii) Stopping potential (cut-off potential) (b) calculate energy of photon of wavelenght `3.31 Å` |
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Answer» (a) Work function: The minimum energy required to make an electron free from the metal is called work function. It is constant for a metal and denote by `phi` or `W`. It is the minimum for cesium. It is relatively less or alkali metals. (ii) Stopping potential: Minimum magnitude of negative potential of anode with respect to cathode for which current is zero is called stopping potential. This is also known as cuttof voltage. This voltage is independent of intensity. b. `E=(12420)/(3.31)eV= 3.752 KeV` |
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| 173. |
Calculate de Broglie wavelength associated with an electron, accelerated through a potential difference of `400 V`. |
| Answer» `lambda=(12.27)/(sqrt(V))=(12.27)/(sqrt(400))=0.613Å` | |
| 174. |
The graph shows variation of stopping potential `V_(0)` versus frequency of incident radiation `v` for two photosensitive metals `A` and `B`. Which of two metals has higher threshold frequency and why? |
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Answer» `K_(max)=hupsilon0-phi` `V=(hupsilon)/(e )-(varphi)/(h)` `varphi_(A) gt varphi_(B)` `upsilon_(A_0) gt upsilon_(B_(0))` |
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| 175. |
In figure find which is `K_(a) and K_(beta)` |
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Answer» `Delta E = (hc)/(lambda), lambda = (hc)/(Delta E)` Since energy different of `K_(alpha)` is less than `K_(beta)` i.e. ,` Delta E_(k _ alpha) lt Delta E_(k _ beta)`, therefore `lambda_(k_alpha) gt lambda_(k_beta)` So , peak `1 is K_(beta)` and peak `2` is `K_(alpha)` |
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| 176. |
consider a hypothetical atom with single electron . In this atom, when an electron de-excites from energy level `n=xx` to `n=2,` wavelength (lambda) of the radiation emitted is given by lambda =`(Ax^(2))/(x^(2)-4)`(where`A` is a eonstant). Choose the correct alternatives.A. Least energetic photon emitted during such a transition will have wavelength `1.8A.`B. Most energetic photon emitted in such trasition will have Wavelength A.C. Ionization potential of the atom in its ground state is `(hc)/(1.8eA) `D. Ionization potential of the atom in its first excited statei (hc)/(eA) |
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Answer» Correct Answer - A::B::D `lambda=(Ax^(2))/(x^(2)-4)Rightarrow1/lambda=4/A(1/2^(2)-1/2^(2))` ` Rightarrow1/lambda_(max)=4/A(1/4-1/9)Rightarrowlambda_(max)=1.8A` For most energetic photon `x=oo Rightarrow 1/lambda_(min)=4/A(1/4-1/oo)Rightarrowlambda_(min)=A` for ionization potential in its ground state `1/lambda=4/A(1/1^(2)-1/oo^(2))Rightarrow lambda=0.25A` ` i.e eV=(hc)/lambda Rightarrow=(hc)/(exx0.25A)` for ionization potential in its first excited state`1/lambda=4/A(1/2^(2)-1/oo^(2)) Rightarrow =lambda=Ai.e eV=(hc)/lambda=RightarrowV=(hc)/(eA)` |
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| 177. |
The size of atomic nucleus is of the order of............ m and size of the atom is of the order of........... .A. `10^(-14)m,10^(-10)m`B. `10^(-10)m,10^(-8)m`C. `10^(-20)m,10^(-16)m`D. `10^(-8)m,10^(-6)m` |
| Answer» Correct Answer - A | |
| 178. |
For electron moving in `n^(th)` orbit of the atom , the angular velocity is proportional to:A. `n`B. `1/n`C. `n^(3)`D. `1/n^(3)` |
| Answer» Correct Answer - D | |
| 179. |
The radius of hydrogen atom , when it is in its second excite state , becomes:A. halfB. doubleC. four timesD. nine times |
| Answer» Correct Answer - D | |
| 180. |
In scattering experiment , the force that scatters particles isA. nuclear forceB. coulomb forceC. Both(1) and (2)D. gravitational Force |
| Answer» Correct Answer - B | |
| 181. |
a-particles areA. helium nucleiB. sodium nucleiC. ionised nucleiD. hydrogen nuclei |
| Answer» Correct Answer - A | |
| 182. |
The energy of electron in first excited state of `H`-atom is `-3.4 eV` its kinetic energy isA. `-3.4 eV`B. `+3.4 eV`C. `-6.8 eV`D. `6.8 eV` |
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Answer» Correct Answer - B Kinetic energy `=|"Total enegry"|` |
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| 183. |
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 ……..A. `1`B. `-1`C. `2`D. `-12` |
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Answer» Correct Answer - B `T.E=-(13.6Z^(2))/n^(2)` `{:(thereforeK.E=T.E),(T.E=-K.E):}` |
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| 184. |
The intensity of an X-ray beam reduces to `36.8%` of its initial intensity after traversing a gold film of thickness `5xx10^(-3)m`. Its absorption coefficient isA. `50 m^(-1)`B. `100 m^(-1)`C. `150 m^(-1)`D. `200 m^(-1)` |
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Answer» Correct Answer - D `I=I_(0)e^(mux)` 0.368 `I_(0)=I_(0)e^(-mux)` `(e^(-1)=0.368)` `mu=1//x =200` |
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| 185. |
This question has statement - 1 and statement - 2 of the four choice given after the statements choose the one that best describes the two statements statement - 1 : A metallic surface is irradiated by a monochromatic light of frequency ` v gt v_(0)` (the threshold frequency). The maximum kinetic energy and the stopping potential are `K_(max) and V_(0)` respectively if the frequency incident on the surface is doubled , both the `K_(max) and V_(0)` are also doubled statement - 2 : The maximum kinetic energy and the stopping potential of photoelectron emitted from a surface are linearly dependent on the frequency of incident light |
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Answer» Correct Answer - D `KE_(max)=eV_(s)=hv-w Rightarrow` v is doubled `KE_(max)` and `V_(s)` become more than doubled and vary linearly with frequency . |
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| 186. |
when a positively charged particle enter a uniform magnetic field with uniform velocity its trajectory can be (a) straight line (b) a cicle a (c)helixA. a onlyB. a or b C. a or cD. any one of a,b and c |
| Answer» Correct Answer - D | |
| 187. |
when a positively charged particle enter a uniform magnetic field with uniform velocity its trajectory can be (a) straight line (b) a cicle a (c)helixA. a onlyB. a or bC. a or cD. any one of a,b and c |
| Answer» Correct Answer - D | |
| 188. |
A Hydrogen atom and `Li^(++) ion` are both in the second excited state . If `l_(H) ` and `l_(Li)` are their respective energies, thenA. `l_(H) gt l_(L I) and |E_(H) | gt |E_(L I)|`B. `l_(H) = l_(L I) and |E_(H) | lt |E_(L I)|`C. `l_(H) = l_(L I) and |E_(H) | gt |E_(L I)|`D. `l_(H) = l_(L I) and |E_(H) | lt |E_(L I)|` |
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Answer» Correct Answer - B `L = (n h)/(2 pi), |E| prop Z^(2)//n^(2) : n = 3` `implies L_(H) = L_(Li) and |E_(H)| lt |E_(LI)|` |
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| 189. |
Consider a metal used to prodeced some charateristic X-ray are give by `E` and wavelength as represented by `lambda`. Then which of the following is true:A. `E(K_(alpha)) gt E(K_(beta)) gt E(K_(gamma))`B. `E(M_(alpha)) gt E(L_(alpha)) gt E(K_(alpha))`C. `lambda(K_(alpha)) gt lambda(K_(beta)) gt lambda(K_(gamma))`D. `lambda(M_(alpha)) gt lambda(L_(alpha)) gt lamda(K_(alpha))` |
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Answer» Correct Answer - C::D `K_(alpha)`: transition from `2 rarr1` Similarly for `K_(beta):3 rarr,K_(gamma):4 rarrL_(alpha):3rarrM_(alpha):4rarr3` Now we can compare energy and `lambda`. |
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| 190. |
Energy levels `A, B, C` of a certain atom corresponding to increasing values of energy i.e., `E_(A) lt E_(B) lt E_(C)`. If `lambda_(1), lambda_(2), lambda_(3)` are the wavelengths of radiations correspnding to the transitions `C` to `B, B` to `A` and `C` to `A` respectively, which o fthe following statements is correct? A. `lambda_(3)=lambda_(1)+lambda_(2)`B. `lambda_(3)=(lambda_(1)lambda_(2))/(lambda_(1)+lambda_(2)`C. `lambda_(1)+lambda_(2)+lambda_(3=0`D. `3lambda_(2)=lambda_(3)+2lambda_(2)` |
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Answer» Correct Answer - B `1/lambda_(3)=1/lambda_(1)+1/lambda_(2) Rightarrowlambda_(3)=?` |
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| 191. |
A `100 eV` electron collides with a stationary helium ion `(He^(+))` in its ground state and excites to a higher level. After the collision , `He^(+)` ion emits two photons in succession with wavelength `1085 Å and 304 Å`. Find the principal quantum number of the excite in its ground state and. Also calculate energy of the electron after the collision. Given `h = 6.63 xx 10^(-34) J s`. |
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Answer» Correct Answer - `5, 47.7 eV` The energy of the electron in the `n^(th)` state of `He^(+)` ion if atomic number `Z` is given by `E_(n) = - (13.6 eV) (Z^(2))/(n^(2))` For `He^(+)` ion, `Z = 2`, therefore `E_(n) = - ((13.6 eV) xx (2)^(2))/(n^(2))` ` = - (54.4)/(n^(2)) eV` The energies `E_(1)` and `E_(2)` of the two emitted photons , in `eV`, are `E_(1) = (12431)/(1085) eV = 11.4 eV` and `E_(2) = (12431)/(304) eV = 40.9 eV` Thus, total energy `E = E_(1) + E_(2) = 11.4 + 40.9 = 52.3 eV` Let `n` be the principal quantum number of the excited state. Using Eq. (i) , we have for the transition from `n = n to n = 1`. `E = - (54.4 eV) ((1)/(l^(2)) - (1)/(n^(2)))` But `E = 52.3 eV`, therefore `52.3 eV = 54.4 eV xx (1 - (1)/(n^(2)))` or `1 - (1)/(n^(2)) = (52.3)/(54.4) = 0.96` `implies n^(2) = 25 or n = 5`. The energy of the electron `= 100 eV` (given). The energy supplied to `He^(+)` ion = `52.3 eV`. Therefore , the energy of the electron left after the collision `= 100 - 52.3 = 47.7 eV` |
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| 192. |
Magnetic field at the center (at nucleus) of the hydrogen like atom `("atomic number" = z)` due to the motion of electron in nth orbit is proporional toA. `(n^(3))/(z^(5))`B. `(n^(4))/(z)`C. `(z^(2))/(n^(3))`D. `(z^(3))/(n^(5))` |
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Answer» Correct Answer - D `B_(n) = (mu_(0)l_(n))/(2 r_(n))` or `B_(n) prop (l_(n))/(r_(n))` `prop ((f_(n)))/(r_(n))` `:. B_(n) prop (v_(n)//r_(n))/((r_n))` `prop (v_(n))/((r_(n))^(2))` `prop ((z//n))/((n^(2)//z)^(2))` `prop (z^(3))/(n^(5))` |
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| 193. |
Magnetic moment due to the motion of the electron in nth energy of hydrogen atom is proportional toA. nB. `n^(0)`C. `n^(5)`D. `n^(3)` |
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Answer» Correct Answer - A `("Magnetic moment")/("Angular momentum") = (e)/(2 m)` :. Magnetic moment prop Angular momentum `prop n ("therefore" L=nh/(2pi))` |
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| 194. |
The wavelength of light emitted from second orbit to first orbits in a hydrogen atom isA. `1.215 xx 10^(-7)m`B. `1.215 xx 10^(-5) m`C. `1.215 xx 10^(-4) m`D. `1.215 xx 10^(-3) m` |
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Answer» Correct Answer - A Energy radiated `E = 10.2 eV = 10.2 xx 1.6 xx 10^(-19) J` `rArr E = (hc)/(lambda) rArr = 1.215 xx 10^(-7) m` |
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| 195. |
Given mass number of gold `= 197`, Density of gold `= 19.7 g cm^(-1)`. The radious of the gold atom is appoximately:A. `1.5 xx 10^(-8) m`B. `1.5 xx 10^(-9) m`C. `1.5 xx 10^(-10) m`D. `1.5 xx 10^(-12) m` |
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Answer» Correct Answer - C Volume occupied by one mole of gold `= (197 g)/(19.7 km ^(-3)) = 10 c m ^(3)` Valume of one atom ` = (10)/(6 xx 10^(-23)) = (5)/(3) xx 10^(23) xx 10^(23) c m ^(3)` Let `r` be the radius of the atom . Therefore. `(4)/(3) pi r ^(3) = (5)/(3) xx 10^(23) or r ~= 1.5 xx 10^(-10) m` |
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| 196. |
Photon of frequency `v` has a momentum associated with it. If `c` is the velocity of light, the momentum is:A. `v//c`B. `hvc`C. `hv//c^(2)`D. `h v//c` |
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Answer» Correct Answer - D `E= pc, hv=pc, p=(hv)/(c )` |
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| 197. |
In figure find which is `K_(a) and K_(beta)` |
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Answer» `DeltaE=(hc)/(lambda), lambda=(hc)/(DeltaE)` since energy difference of `K_(alpha)` is less than `K_(beta)` `1` is `K_(beta)` and `2` is `K_(alpha)` |
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| 198. |
Find the momentum of a `12.0 MeV` photon. |
| Answer» `p=(E )/(c )12 MeV//c` | |
| 199. |
Compare `Z_(1) and Z_(2)` |
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Answer» `:. sqrt(v)sqrt(cR((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))).(Z-b)` If `Z` is greater than `v` will be greater, `lambda` will be less `:. lambda_(1) lt lamda_(2)` `:. Z_(1) gt Z_(2)` |
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The energy ratio of two `k_(alpha)` photons obtined in `x` -ray from two metal targets of atomic numbes, `Z_(1)` and `Z_(2)` is ,A. `Z_(1)/Z_(2)`B. `(Z_(1)/Z_(2))^(2)`C. `((Z_(1)-1)/(Z_(2)-1))^(2)`D. `sqrt(((Z_(1)-1))/((Z_(2)-1))` |
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Answer» Correct Answer - C The energy of `K_(alpha)x`-rays photons is directly proportional to `(Z-1)^(2)`. The energy ratio of two `K_(a)` photons obtained in x-ray from two metal targets of atomic number `Z_(1)a` and `Z_(2)` is `((Z_(1)-1)/(Z_(2)-1))^(2)` |
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