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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.
| 401. |
Suppose that means were available for stripping `28` electrons from `_29 Cu` in vapour of this metel. a. Compute the first three energy level for the remaining electron. b. Find the wavelength of the spectral line of the series for which `n_(1) = 1, n_(1),3,4`. What is the ionization potential for the last electron? |
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Answer» Since `E_(n) prop (Z^(2))/(n^(2))` in a hydrogen-like atom , the energy levels will be `29^(2) = 841` time the corresponding energies for hydrogen. `:. E_(n) = 841 xx ((-13.6)/n^(2)) or E_(1) = - 11.44 ke V` `E_(2) = - 2.86 ke V` `E_(3) = - 1.27 ke V` b. Making use of `Delta E = E_(n) - E_(1)` `implies lambda _(1) = 1.44 Å. lambda _(2) = 1.22 Å, and lambda _(3) = 1.15 Å`. Lonization potential is the potential corresponding to energy `|E_(1)|`. `V_(oo) = (|E_(1)|)/(e) = 11.44 kV` |
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| 402. |
The recoil speed of hydrogen atom after it emits a photon in going from `n = 2 state to n = 1` state is nearly `[Take R_(oo) = 1.1 xx 10^(7) m and h = 6.63 xx 10^(-34) J s]`A. `1.5 m s^(-1)`B. `3.3 m s^(-1)`C. `4.5 m s^(-1)`D. `6.6 m s^(-1)` |
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Answer» Correct Answer - B `E = R_(oo) ch xx [1 - (1)/(2^(2))] = (3)/(4) R_(oo) xx hc` momentum of the photon emitted is, `p = (E)/(c ) = (3 R_(oo) h)/(4)` Recoiling speed of hydrogen atom is given by `v = P//m`. where `m` is the mass of hydrogen atom `v = (3 R_(oo) h)/(4 m) = (3 xx 1.1 xx 10^(7) xx 6.63 xx 10^(-34))/(4 xx 1.67 xx 10^(-27)) = 3.3 m s^(-1)` |
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| 403. |
Consider energy level diagram of a hydrogen atom.How will the kinetc energy and potential energy of electron moves from a lower level to a higher level? |
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Answer» Kinetic energy of an electron in a hydrogen atom is given by `K_(n) = (e^(2))/(8 pi epsilon _(0) r_(n)) = (me^(4))/(8 pi epsilon _(0)^(2) n^(2) h^(2))`, where `K_(0) = (me^(4))/(8 pi epsilon _(0)^(2) h^(2))` Potential energy is given by `U_(n) = (e^(2))/(4 pi epsilon _(0) r_(n)) = (me^(4))/(4 epsilon _(0)^(2) n^(2) h^(2)) = - (2 K_(0))/(n^(2))` The kinetic energy is always positive and it is inversely propertional to square of `n` . Hence , as the electron is excited to higher states , its kinetic energy decreases The potential energy is alyays negative for a partical in bound state. The magnitude of potential energy is twice that of kinetic energy. As the electron is raised to higher energy level,`n` increases. The potential energy becomes less negative that means the potential energy actually increases. Potial energy is maximum for `n rarr oo`, which corresponds to infinite sepaeation between nucleus and electron. |
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| 404. |
The Rutherford `alpha`-particle experiment shown that most of the `alpha`-particles pass through almost unscattered while some are scattered through large angles. What infromation does it given about the structure of the atom ?A. Atom is hollowB. The whole mass of the atom is concentrated in a small centre called nucleusC. Nucleus is positively chargedD. All the above |
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Answer» Correct Answer - D Theory based |
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| 405. |
An alpha particle of energy `5 MeV` is scattered through `180^(@)` by a found uramiam nucleus . The distance of closest approach is of the order ofA. `1Å`B. `10^(-10)cm`C. `10^(-12)cm`D. `10^(-15)cm` |
| Answer» Correct Answer - C | |
| 406. |
An alpha particle of energy `5 MeV` is scattered through `180^(@)` by a found uramiam nucleus . The distance of closest approach is of the order ofA. `1Å`B. `10^(-10)cm`C. `10^(-10)cm`D. `10^(-12)cm` |
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Answer» Correct Answer - C `1/2mv_(0)=(q_(1)q_(2))/(4piepsilon_(0)r)` |
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| 407. |
An alpha particle of energy `5 MeV` is scattered through `180^(@)` by a found uramiam nucleus . The distance of closest approach is of the order ofA. `10^(-15) cm`B. `10^(-13) cm`C. `10^(-12) cm`D. `10^(-19) cm` |
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Answer» Correct Answer - C Use `E = (1)/(4 pi epsilon_(0)) ((Ze) (e))/( r_(0))` |
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| 408. |
In Davosson-Germer experiment, the filament emitsA. photonsB. ProtonsC. X-raysD. Electrons |
| Answer» Correct Answer - D | |
| 409. |
The photoelectrons emitted from a metal surface:A. Are all at restB. Have the same kinetic energyC. Have the same momentumD. Have speeds varying from zero up to certain maximum value |
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Answer» Correct Answer - D Have speeds varying from zero up to a certain maximum value |
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| 410. |
Wrong statement in connection with Davisson-Germer experiment isA. The inter-atomic distance in nickel crystal of the order of the de-broglie wavelength.B. Electrons of constant energy are obtained by the electron gunC. Nickel crystal acts as a three dimensional diffracting grating.D. Davission-Germer experimental is an inteference experiment. |
| Answer» Correct Answer - D | |
| 411. |
In Davisson-Germer experiment the relation between the angle of diffraction `theta` and the grazing angle `0phi` isA. `theta=90^(@)-(phi)/2`B. `phi=90^(@) (theta)/(2)`C. `theta=90^(@)-phi`D. `phi=(90^(@)-theta)//2` |
| Answer» Correct Answer - B | |
| 412. |
Mention the significance of Davisson-Germer experiment. An `alpha`-particle and a phton are accelerated from rest through the same potential difference `V`. Find the ratio of de-Broglie wavelengths associated with them. |
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Answer» Davisson-Germer experiment establishers the existence of matter waves. `K.E`. Gained by a charged particle when accelerated `V` volts is: `K=(1)/(2)mv^(2)=qv` `:. v=sqrt((2qv)/(m))` de-Broglie wavelength of the particle, `lambda=(h)/(mv)=(h)/(msqrt((2qV)/(m)))=(h)/(sqrt(2mqv))` For same `V`, `(lambda(alpha-"particle"))/(lambda("proton"))=sqrt((m_(p)q_(p))/(m_(alpha)q_(alpha)))=sqrt((m_(p)e)/(4m_(p).2e))=(1)/(sqrt(8))=(1)/(2sqrt(2))` |
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| 413. |
An electron microscope uses electrons accelerated by a voltage of `50kV`. Determine the De Broglie wavelength associated with the electrons. If other factors ( such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light? |
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Answer» `lambda = (1.227)/(sqrt(50, 000)) nm`. `5 xx487 xx 10^(-3) nm` Resolving power `prop(1)/(lambda)` [for other factors being fixed] `:. (RP_(em))/(RP_(y,m))=(550 nm)/(5.487xx10^(-3)nm)` `~~(5500xx10^(3))/(5.5)` `~~ 100xx10^(3)` So, `RP` of electron microscope is `10^(5)` times of microscope. Wave length of accelerated electron in electron microscpoe is very small as compared with wavelength of yellow light. |
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| 414. |
A gas of monoatomic hydrogen is bombarded with a atream of electrons that have been accelerated from rest through a potential difference of `12.75 V`. In the emission spectrum, one can observe lines ofA. Lyman seriesB. Balmer seriesC. Paschen seriesD. Pfund series |
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Answer» Correct Answer - A::B::C `E_(n)^(H) = E_(1)^(H) + Delta E = - 13.6 eV + 12.75 eV = - 0.85 eV` , i.e., hydrogen atom are excite to `n = 4` level, i.e., transitions `4 to 1, 3 to 1, 2 to 1` are possible which corresponds to Lyman series, then transitions `4 to 2 and 3 to 2` are possible which corresponds to Balmer series and then transitions `4 to 3` is also possible which correspond to Pachen series. |
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| 415. |
As par Bohr model, the minimum energy (in `eV`) required to remove an electron from the ground state of doubly ionized `Li` atom `(Z = 3)` isA. `1. 51`B. `13.6`C. `40.8`D. `122.4` |
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Answer» Correct Answer - D `E = -Z^(2) xx 13.6 eV =- 9 xx 13.6 eV =-122.4 eV` So ionisation energy `=+ 122.4 eV` |
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| 416. |
As per Bohr model, (I) minimum energy required to remove an electron from ground state of doubly ionized `Li` atom `(Z=3)` is `122.4eV`. (II) energy of transition `n=3` to `m=2` is less than that of `m=2` to `n=1`. (III) minimum energy required to remove an electron from ground state of singly-ionised He atom `(z=2)` is `27.2eV` (IV) A Trasnsition from state `n=3` to `n=2` in a hydrogen atom result in `U-V` radiationA. `II,III,IV`B. `I,II`C. `II,III`D. `I,III,IV` |
| Answer» Correct Answer - B | |
| 417. |
Two electrons starting from rest are accelerated by equal potential difference.A. they will have kinetic energyB. they will have same lineat momentumC. they will have same de Broglie wave lengthD. They will produce X-rays of same minimum wave length when they strikes different different targets. |
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Answer» Correct Answer - A::B::C Kinetic energy of an electron is given by `K=1/2mv^(2)=eV`=same for both electron where v=speed of electron V=potential difference Linear momentum of two electron may be different, in direction but their magnitude i.e `mv=sqrt2mk` will be same . De-broglie wavelength is given by `lambda=h/(mv)`=same for both electrons minimum wavelength of produced `x` rays is given by `lambda_(min)=(hc)/(eV)`=same for both electron . |
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| 418. |
As per Bohr model , the minimum energy (in eV) required to remove electron from the ground state of doubly ioinized `Li` alom `(Z = 3)` isA. 1.51B. 13.6C. 40.8D. 122.4 |
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Answer» Correct Answer - B For hydrogen and hydrogen-like atom , `E_(n) = - 13.6 ((z)^(2))/((n)^(2)) eV` Therefore ground state energy of doubly ionized lithium atom `(z = 3, n = 1)` will be `E_(n) = (- 13.6) ((3)^(2))/((2)^(2)) = - 122.4 eV` Therefore , Lonization energy of an electron in ground state of doubly ionized lithium atom will be `122.4 eV` |
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| 419. |
Consider the spectral line resulting from the transition `n = 2 rarr n = 1 ` in the atoms and ions given . The shortest wavelength is produced byA. hydrogen atomB. deuterium atomC. singly ionization heliumD. doubly ionized lithium |
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Answer» Correct Answer - D We know that `(1)/(lambda) = R z^(2) [(1)/(n_(2)^(2)) - (1)/(n_(1)^(2))] implies (1)/(lambda) prop z^(2)` `lambda` is shortest when `1//lambda` is i.e., when `z` is big `z` is highest for lithum. |
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| 420. |
The `K_(alpha)` X-ray emission line of lungsten accurs at `lambda = 0.021 nm`. What is the energy difference between `K`and `L` levels in the atom?A. `0.51 meV`B. `1.2 meV`C. `59 meV`D. `13.6 meV` |
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Answer» Correct Answer - A `E = (hc)/(lambda) = [(6.63 xx 10^(-34) xx 3 xx 10^(8))/(0.021 xx 10^(-9))] J` `= (6.63 xx 10^(-34) xx 3 xx 10^(8))/(0.021 xx 10^(-9) xx 1.6 xx 10^(-13)) MeV` `= 591.96 xx 10^(-4) MeV = 59.196 keV` |
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| 421. |
The `K_(alpha)` X-ray emission line of lungsten accurs at `lambda = 0.021 nm`. What is the energy difference between `K`and `L` levels in the atom? |
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Answer» `Delta E = (hc)/(lambda) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(0.021 xx 10^(-9))` `= 9.46 xx 10^(-15)J = 5.9 xx 10^(4) e V = 59 ke V` |
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| 422. |
The energy levels of an atom are as shown in figure . Which one of those transition will result in the emission of a photon of wavelength `275 nm`? |
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Answer» Here, `lambda = 275 nm = 275 xx 10^(-9) m` Therefore, energy of the emitting photon, `E = (hc)/(lambda) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(275 xx 10^(-9))` `= 7.22 xx 10^(-19) J = (7.22 xx 10^(-19))/(1.6 xx 10^(-19)) = 4.5 e V` The photon of energy `4.5 e V` ("or of wavelength 275 nm")` will be emitted , corresponding to the transition `B`. |
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| 423. |
When a hydrogen atom is excited from ground state to first excited state, thenA. its kinetic energy increased by `20 eV`B. its kinetic energy dereased by `10.2 eV`C. its potential energy increased by `20.4 eV`D. its angular momentum increased by `1.05 xx 10^(-34) J s` |
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Answer» Correct Answer - B::C::D Ground state `n = 1` First excited state `n = 2` `KE = (1)/(4 pi epsilon_(0)) (e^(2))/(2 r) (z = 1)` `KE = (14.4 xx 10^(-10))/(2 r) eV` Now `r = 0.53 n^(2) Å (z = 1)` `(KE)_(1) = (14.4 xx 10^(-10))/(2 xx 0.53 xx 10^(-10)) eV = 13.58 eV` `(KE)_(2) = (14.4 xx 10^(-10))/(2 xx 0.53 xx 10^(-10) xx 4) eV = 3.39 eV` `KE` decrease by `10.2 eV` Now `PE = (-1)/(4 pi epsilon_(0)) (e^(2))/(r) = (-14.4 xx 10^(-10))/( r) eV` `(PE)_(1) = (-14.4 xx 10^(-10))/(0.53 xx 10^(-10)) eV = - 27.1 eV` `(PE)_(2) = (-14.4 xx 10^(-10))/(0.53 xx 10^(-10) xx 4) eV = - 6.79 eV` `PE` increase by `20.4 eV` Now, angular momentum, `L = mvr= (nh)/(2 pi)` `L_(2) - L_(1) = (nh)/(2 pi) = (6.6 xx 10^(-34))/(6.28) = 1.05 xx 10^(-34) J s` |
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| 424. |
In hydrogen atom, the area enclosed by `n^(th)` orbit is `A_(n)`. The graph between log `(A_(n)/A_(1))` log will beA. B. C. D. |
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Answer» Correct Answer - A `A_(n)propn^(2)` |
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| 425. |
The radius of hydrogen atom in the ground state is `5.3 xx 10^(-11)` m. When struck by an electron, its radius is found to be `21.2 xx 10^(-11) m`. The principal quantum number of the final state will beA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B `r_(n) prop n^(2)` `(n^(2)/(n^(2)) = (21.2 xx 10^(-11))/(5.3 xx 10^(-11)))` or `(n^(2)/(n^(2))) = 4` or `(n^(2)/(1^(2))) = 4` or `n = 2` |
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| 426. |
How many time does the electron go round the first bohr orbit of hydrogen atoms in `1 s`?A. `6.62 xx 10^(15)`B. 100C. 1000D. 1 |
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Answer» Correct Answer - A Frequency of electron revolution: `f = (m Z^(2) e^(4))/(4 epsilon_(0)^(2) n^(3) h^(3))` Put the various value to get `f = 6.62 xx 10^(15) (Z^(2))/(n^(3))` Now put `Z = 1 and n = 1` to get `f = 6.62 xx 10^(15) H z` |
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| 427. |
How many time does the electron go round the first bohr orbit of hydrogen atoms in `1 s`? |
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Answer» Here we have to fine the frequency which is given as `f = (mZ^(2)e^(4))/(4 epsilon_(0)^(3)h^(3))` `= (9.1 xx 10^(-31) xx l^(2) xx (1.6 xx 10^(-19))^(4))/(4 xx (8.85 xx 10^(-12))^(2) xx l^(3) xx (6.6 xx 10^(-34))^(3))` `= 6.62 xx 10^(15)Hz` |
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| 428. |
A hydrogen atom emits a photon corresponding to an electron transition from `n = 5` to `n = 1`. The recoil speed of hydrogen atom is almost (mass of proton `~~1.6 xx 10^(-27) kg)`.A. `10 ms^(-1)`B. `2 xx 10^(-2) ms^(-1)`C. `4 ms^(-1)`D. `8 xx 10^(2) ms^(-1)` |
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Answer» Correct Answer - C The hydrogen atom before the transition was at rest. Therefore from conservation of momentum. `p_((H-atom))=p_("photon") = (E_("radiated"))/(c) = (13.6((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))eV)/(c)` `1.6 xx10^(-27)xxv =(13.6((1)/(1^(2))-(1)/(5^(2)))xx1.6 xx10^(-19))/(3xx10^(8))` `rArrv = 4.352m//s ~~4 m//sec` |
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| 429. |
A hydrogen atom emits a photon corresponding to an electron transition from `n = 5` to `n = 1`. The recoil speed of hydrogen atom is almost (mass of proton `~~1.6 xx 10^(-27) kg)`. |
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Answer» Correct Answer - D `E_("photon")=13.6(1-1/25)eV=13.0eV` ` E//c=mv("momentum conserved")` `E/(mc)=(13xx106xx10^(-19))/(1.67xx10^(-27)xx3xx10^(8))=4m//sec` |
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| 430. |
A hydrogen atom emits a photon corresponding to an electron transition from `n = 5` to `n = 1`. The recoil speed of hydrogen atom is almost (mass of proton `~~1.6 xx 10^(-27) kg)`.A. `10^(-4)m//s`B. `2xx10^(-2)m//s`C. `4m//s`D. `8xx10^(-2)m//s` |
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Answer» Correct Answer - C From law of conservation of momentum `thereforemv=p=h/lambdaRightarrowv=h/(mlambda` |
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| 431. |
If a hydrogen atom at rest, emits a photon of wavelength `lambda`, the recoil speed of the atom of mass m is given by :A. `h/(mlambda)`B. `(mh)/lambda`C. `mhlambda`D. none of these |
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Answer» Correct Answer - A From conservation of linear momentum , after emission of a photon , magnitude of linear momentum of the atom=magnitude of linear momentum of photon i.e `mv=h/lambdaRightarrowv=h/(mlambda)` |
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| 432. |
The intensity of X- ray from a coolidge tube is plotted againest wavelength `lambda` as shown in the figure . The minimum wavelength found `lambda_(c)` and the wavelength of the` k_(a)` line is `lambda_(lambda)`, As the accelerating voliage is increaseA. `lambda_(k)-lambda_(c)` increasesB. `lambda_(k)-lambda_(c)` decreasesC. `lambda_(k)` increasesD. `lambda_(k)` decreases |
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Answer» Correct Answer - A As `lambda_(min)=(hc)/(eV) as V "increase"` lambda_("min")` decrease but the pack of characteristic `x`-rays spectrum remains unchanged. |
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| 433. |
The largest wavelength in the ultraviolet region of the hydrogen spectrum is `122nm`. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest interger) is (a) `802nm` (b) `823nm` (c ) `1882nm` (d) `1648nm`.A. `802nm`B. `823nm`C. `1882nm`D. `1648nm` |
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Answer» Correct Answer - B `1/lambda=PZ^(2) (1/n_(1)^(2)-1/n_(2)^(2))` |
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| 434. |
In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelangth in the Balmer series is:A. `(5)/(27)`B. `(4)/(9)`C. `(9)/(4)`D. `(27)/(5)` |
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Answer» Correct Answer - A For Lyman series `((1)/(lambda_(max)))_(L) = R(1)^(2) [(1)/((1)^(2)) - (1)/((2)^(2))]` `(lambda_(max))_(L) = (4)/(3R)` For Balmer series `((1)/(lambda_(max)))_(B) = R(1)^(2) [(1)/((1)^(2)) - (1)/((3)^(2))]` `(lambda_(max))_(B) = (36)/(5R)` `((lambda_(max))_(L))/((lambda_(max))_(B)) = (4)/(3R) xx (5R)/(36) = (5)/(27)` |
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| 435. |
In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelangth in the Balmer series isA. `5//27`B. `1//93`C. `4//9`D. `3//2` |
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Answer» Correct Answer - A In Lyman series `lambda_(max) = (4)/(3R)` In Balmer series `lambda_(max) = (36)/(5R)`. So required ratio `= (5)/(27)` |
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| 436. |
In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelangth in the Balmer series is:A. `3/23`B. `5/27`C. `7/29`D. `9/31` |
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Answer» Correct Answer - B `1/lambda=RZ^(2)(1/n_(1)^(2)-1/n_(2)^(2))` for longest wave length in Lyman series `n_(1)=1, n_(2)` and for Balmar series `n_(1), n_(2)=3` |
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| 437. |
(a)Find the wavelength of the radiation required to excited the electron is `Li^(++)` from the first to the third Bohr orbit (b) How many spectral lines are observed in the emission spectrum of the above excited system?A. `108.8eV,3`B. `13.6eV,4`C. `54.4eV,2`D. `10.2eV,3` |
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Answer» Correct Answer - A (a) the energy in the first orbit =`E_(1)=Z^(2)E_(0)` Where ltbr. `E_(0)-13.6eV` is the energy of a hydrogen atom in ground state . Thus for `Li^(++)`, ltbr.` E_(1)=9E_(0) =9xx(-13.6eV)`ltbr. The energy in the third orbit is ` E_(3)=E_(1)/n_(2)=E_(1)/9=-13.6eV` Thus , `E_(3)-E_(1) =108.8eV,(hc)/lambda=e_(a)-E_(1)` |
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| 438. |
Let `v_(1)` be the frequency of series limit of Lyman series, `v_(2)` the frequency of the first line of Lyman series and `v_(3)` the frequency of series limit of Balmer series. Then which of the following is correct ?A. `v_(1) - v_(2) = v_(3)`B. `v_(2) - v_(1) = v_(3)`C. `v_(3) = (1)/(2) (v_(1) + v_(2))`D. `v_(1) + v_(2) = v_(3)` |
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Answer» Correct Answer - A Series limit means the shortest possible wavelength (maximum photon energy) and first line means the largest possible wvalength (minimum photon energy) in the series `v = C [(1)/(n^(2)) - (1)/(n^(2))]` (where `C` is a constant) For series limit of Lyman series: `n = 1, m = oo, :. v_(1) = C` For first line of Lyman series: `n = 1, m= 2, :. v_(2) = 3 C//4` For series limit of Balmer series : `n = 2, m = oo :. v_(3) = C//4` |
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| 439. |
Let `v_(1)` be the frequency of series limit of Lyman series, `v_(2)` the frequency of the first line of Lyman series and `v_(3)` the frequency of series limit of Balmer series. Then which of the following is correct ?A. `v_(1) - v_(2) = v_(3)`B. `v_(2) - v_(1) = v_(3)`C. `v_(3) = (1)/(2) ( v_(1) + v_(2)`D. `v_(2) + v_(1) = v_(3)` |
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Answer» Correct Answer - A Series limit means the shortest possible wavelength (maximum photon energy) and first line means the largest possible wavelength (minimum photon energy) in the series. `v = C [(1)/(n^(2)) - (1)/(m^(2))]` (where `C` is a constant) For series limit of Lyman series : `n = 1, m = oo implies v_(1) = C` For first line Lyman series: `n = 1, m = oo implies v_(2) = 3 C//4` For series limit of Balmer series : `n = 2, m = oo implies v_(3) = C//4`. |
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| 440. |
Let `v_(1)` be the frequency of series limit of Lyman series, `v_(2)` the frequency of the first line of Lyman series and `v_(3)` the frequency of series limit of Balmer series. Then which of the following is correct ?A. `vartheta ^(1)-vartheta^(2)=vartheta_(3)`B. `vartheta_(2)-vartheta_(1)=vartheta_(3)`C. `2vartheta_(3)=vartheta_(1)+vartheta_(2)`D. `vartheta-(1)+vartheta_(2)=vartheta_(3)` |
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Answer» Correct Answer - A `1/lambda=R(1/2^(2)-1/a^(2))` series limit for balmer series |
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| 441. |
The energy of electron in the nth orbit of hydrogen atom is expressed as `E_(n) = (-13.6)/(n^(2))eV`. The shortest and longest wavelength of Lyman series will beA. `910 Å, 1213 Å`B. `5463 Å, 7858 Å`C. `1315 Å, 1530 Å`D. None of these |
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Answer» Correct Answer - A `(1)/(lambda_(max)) = R[(1)/((1)^(2)) - (1)/((2)^(2))] rArr lambda_(max) = (4)/(3R) ~~ 1213 Å` and `(1)/(lambda_(min)) = R [(1)/((1)^(2)) - (1)/(oo)] rArr lambda_(min) = (1)/(R ) ~~ 910 Å` |
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| 442. |
Find the longest and shortest wavelengths in the Lyman series for hydrogen. In what region of the electromagnetic spectrum does each series lie? |
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Answer» The transition equation for Lyman series is given by `(1)/(lambda) = R ((1)/(1^(2)) - (1)/(n^(2)))`. N = 2,3` …. The largest wavelength is corresponds to `n = 2` `:. (1)/(lambda_(max)) = 1.097 xx 10^(7)) ((1)/(1) - (1)/(4))` `= 0.823 xx 10^(7)` `implies lambda_(max) = 1.2154 xx 10^(7) m = 1215 Å` The shortest wavelength corresponds to `n = oo` `:. (1)/(lambda_(min)) = 1.097 xx 10^(7)) ((1)/(1) - (1)/(oo))` or ` lambda_(min) = 0.911 xx 10^(-7) m = 911 Å` Both of these wavelength lie in the ultraviolet `(UV)` region of electromagnetic spectrum. S |
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| 443. |
The groud state energy of hydrogen atom is `-13.6 eV`. When its electron is in first excited state, its exciation energy isA. `3.4 eV`B. `6.8 eV`C. 10.2 eV`D. zero |
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Answer» Correct Answer - C Excitation energy is defined as the energy required to take the electron from ground level orbit to any higher order orbit (i.e, `n = 2, 3, 4……)`. Given, ground state energy of hydrogen atom `E_(1)=- 13.6 eV` Enegry of electron in first excited state (i.e., `n = 2)` `E_(2) =- (13.6)/((2)^(2)) eV` Therefore, exciation energy `Delta E = E_(2) - E_(1)` `=- (13.6)/(4) - (-13.6)` `=- 3.4 + 13.6 = 10.2 eV` |
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| 444. |
The total energy of eletcron in the ground state of hydrogen atom is `-13.6 eV`. The kinetic enegry of an electron in the first excited state isA. `3.4` eVB. `6.8` eVC. `13.6` eVD. `1.7` eV |
| Answer» Correct Answer - A | |
| 445. |
The energy of a hydrogen atom in the ground state is `-13.6 eV`. The eneergy of a `He^(+)` ion in the first excited state will beA. `-13.6` eVB. `-27.2` eVC. `-54.5` eVD. `-6.8` eV |
| Answer» Correct Answer - A | |
| 446. |
If the wavelength of the first line of the Balmer series of hydrogen is `6561 Å`, the wavelngth of the second line of the series should beA. 13122ÅB. 3280ÅC. 4860ÅD. 2187Å |
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Answer» Correct Answer - C |
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| 447. |
The minimum wavelength `lambda_min` in the continuous spectrum of X-rays isA. Proportional to the potential difference V between the cathode and anode.B. Inversely proportional to potential differece V between the cathode and anodeC. Proportional to the square root of the potential difference V between the cathode and the anode.D. Inversely proportional to the square root of the potential difference V between the cathode and the anode. |
| Answer» Correct Answer - B | |
| 448. |
Production of continuous X-rays is caused byA. Transition of electrons from higher levels to lower levels in target atomsB. Retardation of incident electron when it enters the target atom.C. Transistor of electrons from lower levels to higher levels in target atoms.D. Neutrasing the incident electron. |
| Answer» Correct Answer - B | |
| 449. |
The wavelength of the first spectral line in the Balmer series of hydrogen atom is `6561 A^(@)` . The wavelength of the second spectral line in the Balmer series of singly - ionized helium atom isA. `1215A^(0)`B. `1640A^(0)`C. `2430A^(0)`D. `4687A^(0)` |
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Answer» Correct Answer - A `1/(6561)=R(1/4-1/9)=(5R)/(36)(therefore1/lambda=RZ^(2)(1/n_(1)^(2)-1/n_(2)^(2)))` ` 1/lambda=4R(1/4-1/16)=(3Rxx4)/16Rightarrowlambda1215A^(0)` |
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| 450. |
which of the following statement about `x`rays is false?A. `E(K_(alpha))+E(L_(beta))+E(M_(alpha))=E(K_(gamma))` where E is the energy of respective `x`-raysB. for the harder `x`-rays the intensity is higher then soft `x`-raysC. the continous and characteristic `x`-rays differ only in the method of creationD. the cut-off wavelength lambda_(min) depend only on the acceleration voltage applied between the target and the filament. |
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Answer» Correct Answer - B Penetrating power of hard `x`-rays is more than soft `x`-rays and penetrating power depends on frequency but not on intensity. |
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