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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.
| 501. |
Kinetic energy of emited cathode rays is dependent onA. only voltageB. only work functionC. both (a) and (b)D. it does not depend upon any physical quantity |
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Answer» Correct Answer - C Higher the voltage, higher is the `K.E.`Higher the work function, smaller is the `K.E.` |
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| 502. |
The extreme wavelength of Paschen series areA. `0.365 mu m` and `0.565 mu m`B. `0.818 mu m` and `1.89 mu m`C. `1.45 mu m` and `4.04 mu m`D. `2.27 mu m` and `7.43 mu m` |
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Answer» Correct Answer - B In Paschen series `(1)/(lambda_(max)) = R [(1)/((3)^(2)) - (1)/((4)^(2))]` `rArr lambda_(max) = (144)/(7R) = (144)/(7 xx 1.1 xx 10^(7))` `= 1.89 xx 10^(-6)m = 1.89 mu m` Similarly `lambda_(min) = (9)/(R ) = (9)/(1.1 xx 10^(7)) = 0.818 mu m` |
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| 503. |
The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation,A. the intensity increaseB. the minimum wavelength increasesC. the intensity remian unchangedD. the mininmum wavelength decreases |
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Answer» Correct Answer - A::B::C As the potential difference is increased , the kinetic energy is incresed. The total energy of `x`-rays emitted will also increase hence intensity will increase. The shorter wavelength will also decrease . |
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| 504. |
A `pi-meason` hydrogen atom is a bound state of negative charged pion (denoted by `pi^(bar), m_(pi) = 273 m_(e))` and a proton. Estimate the number of revolutions a `pi-meason` makes (averagely ) in the ground state on the atom before , it decays (mean life of a `pi-meason ~= 10^(-8) s`, mass of proton `= 1.67 xx 10^(-27) kg)`. |
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Answer» Correct Answer - `1.77 xx 10^(10)` revolutions The frequency of revolution of an electron in hydrogen atom in first orbit is given by `f_(1) = (4 pi^(2) K^(2) e^(4) m)/(h^(2))` Here, in case of a `pi-meson` the mass of electron in replied by the reduced mass as `mu = (m_(p) (273) m)/(m_(p) + 273 m)` Thus the frequency of revolution of `pi-meson` will become `f_(1) = (4 pi^(2) Ke^(4)m_(p) (273) m)/(h^(2)(m_(p) + 273 m))` `4 xx (3.14 xx 9 xx 10^(9) (1.6 xx 10^(-19))^(4)` `= (xx (1.67 xx 10^(-27)) (273 xx 9.1 xx 10^(-31)))/((6.63 xx 10^(-34)) ^(3) (1.67 xx 10^(-27) + 273 xx 9.1 xx 10^(-31)))` `= 1.77 xx 10^(18) s^(-1)` Thus, in the lift of `10^(-8) s`, number of revolation made by the `pi-meson` is given as `N = f_(1) xx Delta t` or `= 1.77 xx 10^(18) xx 10^(-8)` or `= 1.77 xx 10^(10)` revolutions |
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| 505. |
A sample of hydrogen gas in its ground state is irradiated with photons of `10.02 eV` energies. The radiation from the above sample is used to irradiate two other sample of excited ionized `He^(+)` and excited ionized `Li^(2+)`, repectively. Both the ionized samples absorb the incident radiation. How many spectral lines are obtained in the spectra of `Li^(2+)`?A. `10`B. `15`C. `20`D. `17` |
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Answer» Correct Answer - B `Li^(+2)` is excitation from `n=3` to `n=6` so, `N=15` |
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| 506. |
A charge oil drop of charge `q` is falling under gravity with termined velocity v in the absence of electric filed . A electric field can keep the oil drop stationary . If the drop acquires an additional charge , it moves up with velocity `3v` in that field . find the new charge on the drop. |
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Answer» `mg =6pi eta r upsilon, Eq=mg` `Eq^(1) = mg + 6pietar(3upsilon), Eq^(1) = Eq +wq+3Eq` `Eq^(1) =4ERightarrow q^(1) =4q` |
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| 507. |
An electron passes undeflected through perpendicular electric and magnetic filed of intensity `3.4xx10^(3) V//m` and `2xx10^(-3) Wb//m^(2)` respectively . Then its velocity un `m//s` isA. `1.7xx10^(6)`B. `6.8xx10^(6)`C. `6.8`D. `1.7xx10^(8)` |
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Answer» Correct Answer - A `V=E/B` |
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| 508. |
A mono energetic electron beam with a speed of ` 5.2xx10^(6) ms^(-1)` enters into a magnetic filed of induction ` 3xx10^(-4) T`, directed normal to the beam . Find the radius of the circul traced by the beam `(Take e//m=1.76xx10^(11) Ckg)` |
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Answer» From the eqation, `Bev =(mv^(2))/(r)` Radius of circle `=(mv)/(Bq)= V/(B(e//m))` `therefore r=(5.2xx10^(6))/(3xx10^(-4) (1.76xx10^(11)))=01.m` |
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| 509. |
Statement 1: When a beam of highly energetic neutrons is incident on a tungsten target, no X-rays will be produced. Statement 2: Neutrons do not exert any electrostatic force on electrons or nucleus of an atom.A. Statement-1 is true, Statement-2: is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is true, Statement-2: is true, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is true but statement-2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - D Neutron will knock out electrons in target leading to production of characterstic X-ray hence statement-1 is false. |
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| 510. |
When high energetic electron beam , (i.e., cathode rays) strike the heavier metal, then X-ray are produced. Spectrum of X-ray are classified into two categories: (i) continous spectrum, and (ii) characteristic depends only on the potential difference across the electrode. But wavelength of characteristic spectrum depends on the atomic number `(z)`. The production of characteristic X-ray is due to theA. continuous acceleration of incident electrons towards the nucleusB. continuous ratardation of incident electrons towards the nucleusC. electron transition between inner shells of the target atomD. electron transition between outer shells of the target atom |
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Answer» Correct Answer - C Characteristic X-ray series due to transition of electrons. |
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| 511. |
When high energetic electron beam , (i.e., cathode rays) strike the heavier metal, then X-ray are produced. Spectrum of X-ray are classified into two categories: (i) continous spectrum, and (ii) characteristic depends only on the potential difference across the electrode. But wavelength of characteristic spectrum depends on the atomic number `(z)`. The production of continuous X-ray is due to theA. acceleration of incident electrons by the the nucleus of the target atomB. electron transition between inner shells of the target atomC. electron transition between outer shells of the target atomD. annibilation of the mass of incident electrons |
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Answer» Correct Answer - A When the accelerated electron beam strickes the nucleus of target atom, continuous X-ray are produced. |
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| 512. |
Statement-2: When a beam of highly energetic neutrons is incident on a tungsten target, no `X` -rays will be produced. Statement-2: Neutrons do not exert any electrostatics force on electrons or nucleus of an atom |
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Answer» Correct Answer - D Neutrons will knock out electrons in target leading to production opf characteristic `X`-ray. Hence statement-1 is false |
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| 513. |
To produce characteristic `X` - rays using a Tungsten target in an X - ray generator , the accelerating should be greater than ……. Volts and the energy of the characterization is …… eV . (The binding energy of the intermost electron in Tungsten is - 40keV). |
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Answer» Correct Answer - b For minimum accelerating voltage , the electron should jump from `n = 2 to n = 1` level ` For characteristic X-rays , `(1)/(lambda) = R_(alpha) (z - 1) ^(2) [1 - (1)/(n^(2))]` `= R_(alpha) (z - 1) ^(2)` But `E = h ( c)/(lambda) implies (1)/(lambda) = (E)/(hc)` `:. (E)/(hc) = R_(alpha) (E)/(hc)` implies (E_(1))/(E_(2)) = [[1 - (1)/(2^(2))]]/[[1 - (1)/(alpha^(2))]]` ` = E_(1) = (3)/(4) E_(2) = (3)/(4) 40.000 eV = 30.000 eV` |
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| 514. |
When electron jumps from n = 4 to n = 1 orbit, we getA. second line of Lyman seriesB. second line of Balmer seriesC. second line of Paschen seriesD. an absorption line of Balmer series |
| Answer» Correct Answer - B | |
| 515. |
A doubly ionized lithium atom is hydrogen like with atomic number 3. Find the wavelength of the radiation required to excite the electron in `Li^(++)` from to the third Bohr orbit (ionization energy of the hydrogen atom equals 13.6 eV). |
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Answer» The energy of `n^(th)` orbit of a hydrogen -like atom is given as` E_(n) =-(13.6Z^(2))/n^(2)` Thus for `Li^(2+)` atom , as `Z= 3`, the electron energies of the first and third Bohr orbits are For `n =1,E_(1) =- 122.4eV`, for `n=3`, an elecron from `E_(1)` level to `E_(3)` level to `E-(3)` level is `E=E_(3) -E_(1)` ` =-13.6-(-122.4)=108.8e`. Therefore, the radiation needed to cause this transition should have photons of this energy. `hv = 108.8 eV`. The wavelengtth of this radiation is or `=lambda=(hc)/(108.8eV)= 114.25Å` |
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| 516. |
Which of the following transitions in a hydrogen atom emits photon of the highest frequency ?A. n = 1 to n = 2B. n = 2 to n = 1C. n = 2 to n = 6D. n = 6 to n = 2 |
| Answer» Correct Answer - B | |
| 517. |
Which of the following transitions in a hydrogen atom emits photon of the highest frequency ?A. n=1 to n=2B. n=2 to n=1C. n=2 to n=6D. n=6 to n=2 |
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Answer» Correct Answer - A |
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| 518. |
A double ionised lithium atom is hydrogen like with atomic number `3` (i)Find the wavelength of the radiation to excite the electron in `Li^(++)`from the first to the third bohr orbit (lonisation energy of the hydrogen atom equals `13. 6eV` (ii) How many spectral lines are observed in the emission spectrum of the above excited system ? |
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Answer» Correct Answer - `[(a) 113.7Å (b)3]` |
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| 519. |
A caesium photocell , with a steady potential difference of `60 V` across , is alluminated by a bright point source of light `50 cm` away. When the same light is placed `1 m` away the photoelectrons emitted from the cellA. each caryy one quarter of their previous energyB. each carry one quarter of their previous momentumC. are half as numerousD. are one quarter as numerous |
| Answer» Correct Answer - D | |
| 520. |
The energy levels of the hydrogen spectrum is shown in figure. There are some transitions `A, B, C, D` and `E`. Transition `A, B`, and `C` respectively represent A. First member of Lyman series, third spectral line of Balmar series and the second spectral line of Paschen seriesB. Ionization potential of hydrogen, second special line of Balmer series and third spectral line of Paschen seriesC. Series limit of Lyman series, third spectral line of Balmer series and second spectral line of Paschen seriesD. Series limit of Lyman series, second spectral line of Balmer series and third spectral line of Paschen series |
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Answer» Correct Answer - C |
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| 521. |
A double charged lithium atom is equivalent to hydrogen whose atomic numer is `3`. The wavelength of required radiation for emitted electron from first to third Bohr orbit in `Li^(++` will be (lonisation energy of hydrogen atom is `13.6 eV`)A. 182.24 ÅB. 177.17 ÅC. 142.25 ÅD. 113.74 Å |
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Answer» Correct Answer - D |
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| 522. |
The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy `6 eV` fall on it is `4 eV` . The stopping potential , in volt isA. `2V`B. `4V`C. `6V`D. `10V` |
| Answer» Correct Answer - B | |
| 523. |
The stopping potential in an experiment on a photo electric effect is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted? |
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Answer» The maximum kinetic energy of an electron is given by `K_(max)=eV " "[ because V_(0)=1.5V]` `=1.6xx10^(-19)xx1.5=2.4xx10^(-19)J`. |
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| 524. |
In an experiment on photo electric emission, following observations were made, (i) Wavelength of the incident light `=1.98xx10^(-7)m`, (ii) Stopping potential `=2.5 "volt"`. Find: (a) Kinetic energy of photoelectrons with maximum speed. (b) Work function and (c )Threshold frequency, |
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Answer» (a) Since `V_(s)=2.5V`, `K_(max)=eV_(s)` so, `K_(max)=2.5 eV` (b) Energy of incident photon `E=(12400)/(1980)eV=6.26 eV` , `W=E-K_(max)=3.76 eV` (c ) `hv_(th)=W=3.76xx1.6xx10^(-19)J :. v_(th)=(3.76xx1.6xx10^(-19))/(6.6xx10^(-34))~~9.1xx10^(14)Hz` |
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| 525. |
If an X-ray tube operates at the voltage of 10kV, find the ratio of the de-broglie wavelength of the incident electrons to the shortest wavelength of X-ray producted. The specific charge of electron is `1.8xx10^11 C/(kg)`.A. 1B. 0.1C. 1.8D. 1.2 |
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Answer» Correct Answer - B Kinetic energy gained by a charge `q` after being accelerated through a potential difference `V` volt, is given by `q V = (1)/(2) m v^(2)` `m V = sqrt(2 M q V)` As `lambda_(b) = (h)/(m V) = (h)/(sqrt (2 q m V)` for cut- off wavelength of X-ray , we have `q V = (h c)/(lambda_(m))` or `lambda_(m) = (hc)/(q V)` Now, `(lambda_(b))/(lambda_(m)) = sqrt ((qV)/(2 m))/(c)` As `(q)/(m) xx 1.8 xx10^(11) C kg^(-1)` for electron, we have `(lambda_(b))/(lambda_(m)) = sqrt(1.8 xx 10^(11) xx 10 xx 10^(3)//2)/(3 xx 10^(8)) = 0.1` |
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| 526. |
The shortest wavelength produced in an X-ray tube operating at `0.5` million volt isA. dependent on the target elementB. about `2.5 xx 10^(-12) m`C. double of the shortest wavelength produced at `1` milllion voltD. dependent only on the target material |
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Answer» Correct Answer - C `lambda_(min) = (hc)/(eV_(max))` |
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| 527. |
If `lambda_(min)` is minimum wavelength produced in X-ray tube and `lamda_(kalpha)` is the wavelength of `K_(alpha)` line. As the operating tube voltage is increased.A. `(lambda_(k)-lambda_(min))` increasesB. `(lambda_(k)-lambda_(min))` decreasesC. `lambda_(kalpha)` increasesD. `lambda_(kalpha)` decreases |
| Answer» Correct Answer - A | |
| 528. |
If the short wavelength limit of the continous spectrum coming out of a Coolidge tube is `10 A`, then the de Broglie wavelength of the electrons reaching the target netal in the Coolidge tube is approximatelyA. 0.3 ÅB. 3ÅC. 30ÅD. 10Å |
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Answer» Correct Answer - A We have K.E.`=(P^(2))/(2m_(0)) =(hc)/(lambda_("min"))` `rArr P=sqrt((2hcm_(0))/(lambda_("min")))` Also `lambda_("de broglie")=h/p=sqrt((hlambda_("min"))/(2m_(0)C))` For `lambda_("min")=10Å: lambda_("de broglie")=0.3 Å 0` |
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| 529. |
Statement -1: When applied potential difference Between cathode and anode is increased in a cooliedge tube , cute off wavelength decreases. Statement-2: Cut off wavelength for `X`-rays depend on atomic number of target material. |
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Answer» Correct Answer - C The cut off wavelength for `x`-ray depend on applied voltage between cathod and anode and not on atomic number of target. Hence statement -2 is false |
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| 530. |
The wavelength of `K_(alpha)` line for an element of atomic number `43 is lambda`. Then the wavelength of `K_(alpha)` line for an element of atomic number `29` isA. `(43)/(29) lambda`B. `(42)/(28) lambda`C. `(9)/(4) lambda`D. `(4)/(9) lambda` |
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Answer» Correct Answer - C |
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| 531. |
If the potential difference between the anode and cathode of the `X` - ray tube is increases A. the peaks at R and S would move to shorter wavelengthB. the peaks at R and S would remain at the same wavelengthC. the cut- off wavelength at P would decreaseD. (b) and `(c)` both are correct |
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Answer» Correct Answer - D |
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| 532. |
An X-ray tube operates at `20k V`. Find the maximum speed of the electron strinking the anode , given the charge of electron is `1.6 xx 10^(-19)` coulomb and mass of electron is `9 xx 10^(-31) kg`. |
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Answer» When an electron of charge `e` is accelerated through a potential different `V`, it acquired energy `e V`. If `m` be the mass of the electron and `v_(max)` the maximum speed of electron , then `(1)/(2) m v_(max)^(2) = e V` or `v_(max) = sqrt(((2 e V)/(m)))` Subsitituting the given value , we get `v_(max) = sqrt(((2 xx (1.6 xx 10^(-19)) xx 20.000)/(9 xx 10^(-31))))` `= 8.4 xx 10^(7) ms^(-1)` |
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| 533. |
When the number of electron striking the anode of an X - ray tube is increase , the ……. of the emitted X - ray increases , while when the speed of the electrons the anode are increased , the cut - off wavelength of the emitted X - ray ………. |
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Answer» Correct Answer - a Intensity decreases, More the number of electrons striking the anode, more is the emitted numbet of photon of X-ray. `[Q lambda = (12375)/(V) Å]`, where `V` is not volt. |
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| 534. |
Can a hydrogen atom emit characteristic X-ray?A. its energy levels are too close to each otherB. its energy levels are too for apartC. it has a very small massD. it has a single electron |
| Answer» Correct Answer - A | |
| 535. |
The wavelength of the characteristic X - ray `k_(alpha )` line emitted by a hydrogens like element is `0.32 lambda ` . The wavelength of the `K_(beta)` line emitted by the same element will be ….. |
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Answer» Correct Answer - d For `K_(alpha), (1)/(lambda) = c [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` `implies (1)/(0.32) = c [(1)/(1^(2)) - (1)/(2^(2))]` = `(3c)/(4)` For `K_(alpha) = 1 = c ((1)/(1^(2)) - (1)/(3^(2))) = (8c)/(9)` On dividing, we get `lambda = 0.27` |
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| 536. |
Which one of the following statements is wrong in the context of X-ray generated from and X-ray tube?(a) Wavelength of characteristic X-ray decreases when the atomic number of the target increases (b) Cut off wavelength of the continuous X-rays depends on the atomic number of the target (c ) intensity of the characteristic X-rays depends on the electrical power given to the x-ray tube (d) cut off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tubeA. Wavelength of characterstics X-ray decreases when the atomic number of the target increasesB. Cut-off wavelength of the continuous X-rays depends on the atomic number of the targetC. Intensity of the characteristic X-rays depends on the elecreical power given to the X-ray tubeD. Cut-off wavelength of the continuous X-ray depends on the energy of the electrons in the X-ray tube |
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Answer» Correct Answer - B `lambda_(min)=(hc)/(eV)` Cut off wavelength depends on the energy of the accelerated electrons and is independent of nature of target. `lambda_(kalpha)prop(1)/((z-b)^(2))` characterstic wavelength depend on atomic no and cut off wavelength depend on energy of `e^(-)` |
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| 537. |
The `K_(alpha)` `X` - rays arising from a cobalt `(z = 27)` target have a wavelength of `179 pm`. The `K_(alpha)` `X` - rays arising from a nickel target `(z = 28)` isA. `gt179"pm"`B. `lt179 "pm"`C. `=179 "pm"`D. None of these |
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Answer» Correct Answer - B |
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| 538. |
For characteristic `X` - ray of some materialA. `E(K_(gamma))ltE(K_(beta))ltE(K_(alpha))`B. `E(K_(alpha))ltE(L_(alpha))ltE(M_(alpha))`C. `lambda(K_(lambda))ltE(K_(beta))ltE(K_(alpha))`D. `lambda(M_(gamma))ltlambda(L_(beta))ltgamma(K_(alpha))` |
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Answer» Correct Answer - C |
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| 539. |
Assertion (A), study of discharge of electricity through gases at low pressure resulted in the discovery of cathod rays . Reason (R ): cathod rays are deflected by both magnetic and electric fields and the direction of deflection shows that they are nagatievly charged .A. `A` and `R` are true and `R` is the correct explaination of `A`B. `A` and `R` are true and `R`is not the correct explaination of `A`C. `A` is true and `R` is false.D. `A` is false but `R` is true |
| Answer» Correct Answer - B | |
| 540. |
The ratio of moment of an electron and an `alpha`-particle which are accelerated from rest by a potential difference of `100V` isA. `1`B. `sqrt((2m_(e))/(m_(alpha)))`C. `sqrt((m_(e))/(m_(alpha)))`D. `sqrt((m_(e))/(2m_(alpha)))` |
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Answer» Correct Answer - D Momentum `p = mv` and `v = sqrt((2QV)/(m))` |
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| 541. |
Energy of the electron in `nth` orbit of hydrogen atom is given by `E_(n) =-(13.6)/(n^(2))eV`. The amount of energy needed to transfer electron from first orbit to third orbit isA. `13.6 eV`B. `3.4 eV`C. `12.09 eV`D. `1.51 eV` |
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Answer» Correct Answer - C For `n = 1,E_(1) =- (13.6)/((1)^(2)) =- 13.6 eV` and for `n=3, E_(3) =- (13.6)/((3)^(2)) =- 1.51 eV` So required enegry `= E_(3) - E_(1) =- 1.51 -(-13.6) = 12.09 eV` |
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| 542. |
Energy of the electron in `nth` orbit of hydrogen atom is given by `E_(n) =-(13.6)/(n^(2))eV`. The amount of energy needed to transfer electron from first orbit to third orbit isA. `13.6 eV`B. `3.4 eV`C. `12.09 eV`D. 1.51 eV` |
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Answer» Correct Answer - C For `n = 1, E_(1) = - (13.6)/((1)^(2)) = - 1.51 eV` and for `n = 3, E_(3) = - (13.6)/((3)^(2)) = - 1.51 eV` So required energy `= E_(3) - E_(!) = - 1.51 - (-13.6) = 12.09 eV` |
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| 543. |
A gas of hydrogen like ions is prepared in a particular excited state A if emit photons havingwavelength equal to the wavelength of the first line of the lyman series togather with photons of five other wavelength identify the gas and find the principal quantum number of the stateA` |
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Answer» Correct Answer - `[He^(+),4]` |
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| 544. |
Find the temperature at which the everage thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state `n = 3` state hydrogen can now emit rod light of wavelength `653.1 nm`because of maxwellan distribution of speeds a hydrogen sample emits red light at temperature much lower than that obtained from this problem Asuume that hydrogen that hydrogen molecules dissociate into atoms |
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Answer» Correct Answer - `[9.4 xx 10^(4)K]` |
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| 545. |
A hydrogen atom in the normal state is located at a distance `r = 2.5 cm` from a long straight conductor carrying a current `I = 10A`. Find the maximum force acting on the atom. |
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Answer» Correct Answer - `[F = 3 xx 10^(-26)N]` |
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| 546. |
A spectroscopic instrument can resolve two nearly wavelength `lambda and lambda = Delta lambda if lambda//Delta lambda` is smaller than `8000` This is used to study the spectral lines of the belmar series of hydrogen Approxmately how many lines will be resolved by the instrument? |
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Answer» Correct Answer - [38] |
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| 547. |
An `alpha`- particle and a proton are accelerated from rest by a potential difference of `100 V` After this their de Broglie wavelength are `lambda_(a) and lambda_(p)` respectively The ratio `(lambda_(p))/ lambda_(p)` , to the nearest integer is |
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Answer» Correct Answer - `3` `P_(1) = sqrt(2m(100 eV))` `lambda_(P) = (h)/(sqrt(2m(100eV))) rArr lambda_(alpha)=(h)/(sqrt(2(4m)2(100 eV)))` `lambda_(P)/lambda_(alpha)=sqrt(8)` rArr The ratio `lambda_(p)/lambda_(alpha)`, to the nearest integer, is equal to `3`. |
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| 548. |
A proton and an alpha - particle are accelerated through same potential difference. Then, the ratio of de-Broglie wavelength of proton and alpha-particle isA. 2B. `2sqrt(2)`C. `1//2sqrt(2)`D. `sqrt(2)` |
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Answer» Correct Answer - B `I=h/(sqrt(2mqv))` |
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| 549. |
A beam of white light is incident normally on a plane surface absorbing `70%` of the light and reflecting the rest. If the incident beam carries `30 W` of power, find the force exerted by it on the surface. |
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Answer» Correct Answer - `(1.3xx30)/(3xx10^(8))=1.3xx10^(-7)` change in momentum `DeltaP=0.7(h)/(lambda)+2xx0.3xx(h)/(lambda)` `=1.3(P)/(c )=1.3xx(30)/(3xx10^(8))=1.3xx10^(-7)N`. |
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| 550. |
The radiation force experienced by body exposed to radiation of intensity `I`, assuming surface of body to be perfectly absorbing is: A. `(piR^(2)I)/(c )`B. `(piRHI)/(c )`C. `(IRH)/(2 c)`D. `(IRH)/(c )` |
| Answer» Correct Answer - D | |