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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. |
Find the velocity (in `m s^(-1)`) of electron in first Bohr orbit of radius `a_(0)`. Also find the de Broglie wavelength (in m). Find the orbital angular momentum of 2p orbital of hydrogen atom in units of `h//2pi`. |
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Answer» For H and H-like particles, velocity in the nth orbit `v_(n) = 2.188 xx 10^(6) xx (Z)/(n) ms^(-1)` For H-atom, Z = 1 and for 1st orbit n = 1 `:. v = 2.188 xx 10^(6) ms^(-1)` de Broglie wavelength, `lamda = (h)/(mv) = (6.626 xx 10^(-34) kg m^(2) s^(-1))/(9.1 xx 10^(-31) kg xx 2.188 xx 10^(6) ms^(-1)) = 3.33 xx 10^(-10) m` Orbital angular momentum `= sqrt(l(l + 1)) (h)/(2pi)` For 2p orbital, `l = 1` `:.` Orbital angular momentum `= sqrt(1(l + 1)) (h)/(2pi) = sqrt2 (h)/(2pi)` |
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| 502. |
Calculate the wavelength associated with an electron moving with a velocity of `10^(3) m s^(-1)`. |
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Answer» Correct Answer - `7.28xx10^(-7)m` According to de Broglie equation ,`lambda=(h)/(m upsilon)` `h=6626xx10^(-34)"kg "m^(2)s^(-1), m=9.1xx10^(-31)"kg", upsilon=10^(3)m s^(-1)` `:.` Wavelength `(lambda)=((6.626xx10^(-34)"kg"m^(2)s^(_1)))/((9.1xx10^(-31)"kg")xx(10^(3)ms^(-1)))=7.28xx10^(-7)m`. |
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| 503. |
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength `6800 Å`. Calculate the threshold frequency and work function `(W_(0))` of the metal. |
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Answer» Threshold wavelength `(lambfa_(0))=6800 Å=6800xx10^(-10) m` As `c=v lambda :. v_(0)=c/lambda_(0)=(3.0xx10^(8)" ms"^(-1))/(6800xx10^(-10) m)=4.41xx10^(14) s^(-1)` Work function `(W_(0))=hv_(0)=(6.626xx10^(-34) J s)(4.41xx10^(14) s^(-1))=2.92xx10^(-19) J` |
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| 504. |
Calculate the threshold freqency of the metalfrom the photoelectronsare emitted with zero velocity when exposed to radiation of wavelength 6800 `Å` Hint `: KE = (1)/(2) mv^(2)` `hv= hv_(0) + KE, KE=0` `v= (c ) /( lambda)` |
| Answer» `4.41 xx 10^(14) s^(-1)` | |
| 505. |
The effect of uncertainty principle is significant only for motion of microscopic particles and is negligible for the macroscopic particles. Justify the statement with the help of a suitable example. |
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Answer» If uncertainty principle is applied to an object of mass, say about a milligram `(10^(-6)kg)`, then `Delta.Deltax=(h)/(4pim)` `Deltav.Deltax=(6.626xx10^(-34)kgm^(2)s^(-1))` `=0.52xx10^(-28)m^(2)s^(-1)` The value of `Deltav.Deltax` obtianed is extermeely small and is insignificant. Therefore, for miligram-sized or heavier objects, the associated uncertainties are hardly of any real consequence. |
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| 506. |
The work function `(phi)` of some metals is listed below. The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal is `{:("Metal",Li,Na,K,Mg,Cu,Ag,Fe,Pt,W),(phi(eV),2.4,2.3,2.2,3.7,4.8,4.3,4.7,6.3,4.75):}` |
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Answer» Correct Answer - 4 Energy associated with the incident photon `= (hc)/(lamda)` i.e., `E = ((6.6 xx 10^(-34) Js) (3 xx 10^(8) ms^(-1)))/((300 xx 19^(-9) m))` `= 6.6 xx 10^(-19) J = (6.6 xx 10^(-19))/(1.6 xx 10^(-19)) eV = 4.12 eV` `:.` Metals showing photoelectric effect will be Li, Na, K and Mg only i.e., 4 metals, (which have work function less than 4.12 eV) |
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| 507. |
The orbital angular momentum for an electron revolving in an orbit is given by `sqrt(l(l + 1))(h)/(2pi)`. What is the momentum of an s-electron?A. `h//2 pi`B. `sqrt(2) h// pi`C. `+(1)/(2) h// 2pi`D. zero. |
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Answer» Correct Answer - D Momentum for s-electron `=sqrt(0(0+1)) , h//2pi =zero`. |
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| 508. |
Which of the following is given the credit for discovering the electron?A. William CrookesB. G.J.StoneyC. J.J. ThomosnD. Michael Faraday |
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Answer» Correct Answer - C Because he was the first to measure one of its properties, the ratio of the charged of the electron to its mass, represented as `e//m`. |
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| 509. |
The group having isoelectronic species isA. `O^(2-) , F^(-), Na, Mg^(2+)`B. `O^(-), F^(-), Na^(+), Mg^(+)`C. `O^(2-), F^(-), Na^(+), Mg^(2+)`D. `O^(-), F^(-), Na, Mg` |
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Answer» Correct Answer - C `O^(2-) = 8 + 2 = 10 e^(-)1`, `Na^(+) = 11 - 1 = 10 e^(-), Mg^(2+) = 12 - 2 = 10 e^(-)` Hence, group (c) contains isoelectronic species. |
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| 510. |
The series of lines present in the visible region of the hydrogen spectrum isA. LymanB. BalmerC. PaschenD. Brackett |
| Answer» Correct Answer - B | |
| 511. |
What is the energy in joules required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state ? The ground state energy is `-2.18xx10^(-11)` ergs. |
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Answer» As ground state electronic energy is `-2.18xx10^(-11)` ergs, this means that `E_(n)=-(2.18xx10^(-11))/n^(2)` ergs `DeltaE=E_(5)-E_(1)=2.18xx10^(-11) (1/1^(2)-1/5^(2))=2.18xx10^(-11) (24/25)=2.09xx10^(-11)` ergs `=2.09xx10^(-18)J (1 erg =10^(-7) J)` When electron returns to ground state (i.e., to `n=1`), energy emitted `=2.09xx10^(-11)` ergs. As `E=hv=hc/lambda or lambda=(hc)/E=((6.626xx10^(-27)" erg sec")(3xx10^(10)" cm s"^(-1)))/(2.09xx10^(-11) ergs)` `=9.51xx10^(-6) cm=951xx10^(-8) cm=951 Å` |
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| 512. |
The____of lines are the only linesin the hydrogen spectrum which appear in the visible region of the electromagnetic spectrum.A. Pfund seriesB. Brackett seriesC. Paschen seriesD. Balmer series |
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Answer» Correct Answer - D The visible portion of the line spectrum of the `H` atom consists of only four lines. |
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| 513. |
Which of the following statements about the electron is incorrect?A. It is a negatively charged particle.B. The mass of electron is equal to the mass of neutron.C. It is a basic constituent of all atoms.D. It is a constituent of cathode rays. |
| Answer» Correct Answer - B | |
| 514. |
Assertion : When an iron rod is heated in a furnace, the radiation emitted goes from a lower frequency to a higher frequency as the temperature increases. Reason : The energy of a quantum of radiation is proportional to its frequency.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 515. |
Which of the folowing statements is incorrect about an atomic orbital?A. It is a single electron wave function.B. It describes the trajectroy of electron in an atom.C. It define the distribution of electron density in space.D. It can be represented by boundary surface diargams. |
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Answer» Correct Answer - B An orbital is a well defined path (trajectory) around the nucleus in which the eletron moves whereas an orbital is a three-dimensional space around the nucleus within which the total probability of finding the eletorn is around `90%`. |
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| 516. |
The sodium lamp has some advantage over the mercury lamp such as (i) for a given input of electricity, the sodium lamp gives off greater light intensity than does the mercury lamp. (ii) yellow light has a longer wavelength than bluish-green light. (iii) human eye exhibits its greatest reponse for yellow color.A. `(i),(iii)`B. `(ii),(iii)`C. `(i),(ii),(iii)`D. `(i),(ii)` |
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Answer» Correct Answer - C Due to longer wavelength, yellow light is not scatered as much by small particles is the air-fog, for example, and penetrates farther than white light. Moreover, you many have noticed that some autmobile fog lights have yellowtinted glass covers. |
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| 517. |
The electron energy in hydrogen atom is given by `E_(n)=(-2.18xx10^(-18))//n^(2)J`. Calculate the energy required to remove the electron completely from `n=2` orbit. What is the longest wavelength of light in cm that can be used to cause this transition ? |
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Answer» `Delta E=E_(oo)-E_(2)=0-(-(2.18xx10^(-18)" J atom"^(-1))/2^(2))=5.45xx10^(-19)" J atom"^(-1)` `DeltaE=hv=h c/lambda` or `lambda=(hc)/(Delta E)=((6.626xx10^(-34)J s)xx(3xx10^(8) ms^(-1)))/(5.45xx10^(-19) J)=3.647xx10^(-7)m=3.647xx10^(-7)m=3.647xx10^(-5) cm` |
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| 518. |
Which one is a set of isolectronic species ?A. `N_(2), CO_(2), CN^(-)`B. `N, H_(2)S, CO`C. `N_(2), CO, CN^(-)`D. `Ca, Mg, Cl` |
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Answer» Correct Answer - C `N_(2) = 7 + 7 = 14e^(-), CO_(2) = 6 + 2 xx 8 = 22 e^(-)` `CN^(-) = 6 + 7 + 1 = 14 e^(-)`. Hence, (a) is not `N = 7e^(-), H_(2)S = 2 + 16 = 18e^(-)` `CO = 6 + 8 = 14 e^(-)`. Hence (b) is not `N_(2) = 14 e^(-), CO = 6 + 8 = 14e^(-), CN^(-) = 6 + 7 + 1` `= 14e^(-)`. Hence , (c) is a set isolectronic species |
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| 519. |
When difference elements are present in the discharged tube positive ions with different_______are observed. (i) charges (ii) masses (iii) `e//m` ratiosA. `(i),(iii)`B. `(ii),(iii)`C. `(i),(ii),(iii)`D. `(i),(ii)` |
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Answer» Correct Answer - C Bacause different elements give different positive ions. |
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| 520. |
Assertion : In electromagnetic spectrum, the small portion around `10^(15)`Hz is called visible light. Reason :Visible region is only a small part of the entire spectrum which our eyes can see.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 521. |
Which of the following types of spectrum is best depicted by the given figure? A. Atomic absorption spectraB. Atomic emission spectraC. Continuous spectraD. None of these |
| Answer» Correct Answer - B | |
| 522. |
The spectrum of while light ranging from red to violet is called a continuous spectrum becauseA. different colours are seen as different bands in the spectrumB. the colours continuously absorb energy to form a spectrumC. the violet colour merges into blue, blue into green, green into yellow and so onD. it is a continuous band of coloured and white light separating them. |
| Answer» Correct Answer - C | |
| 523. |
Out of the folowing paris of electorns, identify the pairs of electrons present in degenrate orbitals.A. in`=3,l=2,m_(1)=-2,m_(s)=-(1)/(2) iin=3,l=2,m_(1)=-1,m_(s)=-(1)/(2)`B. `in=3,l=1,m_(1)=1,m_(s)=+(1)/(2) iin=3,l=2,m_(1)=1,=+(1)/(2)`C. `i n=4,l=1,m_(1)=1,m_(s)=+(1)/(2) ii n=3,l=2,m_(1)=1,m_(s)=+(1)/(2)`D. `in=3,l=2,m_(1)=+2,m_(s)=-(1)/(2) ii n=3,l=2,m_(1)=+2 ,m_(s)=+(1)/(2)` |
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Answer» Degenerate orbitals means the orbitals of the same subshell of the same main shell, i.e., their n and l value (a) i `3d_(xy)` (ii) `3d(yz)` (b) (i) `3p_( x)` (ii) `3d_(xy)` (c) (i) `4s` (ii) `3d_(x y)` (d) (i) `ed_(x^(2)-y^(2)` (ii) `3d_(x^(2)-y^(2)` Thus `3d_(xy)` and `3d_(yz):3d_(x^(2)-y^(2)` and `3d_(x^(2)-y^(2))` represent pair of degonerate orbitals. |
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| 524. |
In which of the following pairs, the ions are isoelectronic ?A. `Na^(+), Mg^^(2+)`B. `Al^(3+), O^(-)`C. `Na^(+), O^(2-)`D. `N^(3), Cl^(-)` |
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Answer» Correct Answer - A::C `Na^(+) = 11 -1 = 10 e^(-), Mg^(2+) = 12 -2 = 10 e^(-)`. Thus, they have same no. of electrons `Al^(3+) = 13 -3 = 10 e^(-), Q^(2-) = 8 + 2 = 10 e^(-)`. They have same no. of electrons `N^(3-) = 7 + 3 = 10 e^(-), Cl^(-) = 17 + 1 = 18 e^(-)`. They do not have same no. of electrons |
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| 525. |
Among the following groupings which represents the collection of isoelectronic species ?A. `NO^(+), C_(2)^(2-), O_(2)^(-), CO`B. `N_(2), C_(2)^(2), CO, NO`C. `CO, NO^(+), CN^(-), C_(2)^(2)`D. `NO, CN^(-), N_(2), O_(2)^(-)` |
| Answer» Correct Answer - C | |
| 526. |
In which one of the following pairs, the two species are both isoelectronic and isotopic ? (Atomic number : `Ca = 20, Ar = 18, K = 19, Mg = 12, Fe = 26, Na= 11`)A. `.^(40)Ca^(2+) and .^(40)Ar`B. `._(39)K^(+) and .^(40)K^(+)`C. `.^(24)Mg^(2+) and .^(25)Mg`D. `.^(23)Na and .^(24)Na^(+)` |
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Answer» Correct Answer - B In (b), `K^(+)` has 18 electrons in each. As they have same atomic number but different masses, so they are also isotopes |
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| 527. |
Name three ions which are isoelectronic with `F^(-)` ion |
| Answer» Electrons in `F^(-)` ion `= 9 + 1 = 10`. Isoelectronic ions will be `Na^(+), Al^(3+) and O^(2-)` | |
| 528. |
Line spectrum is the characteristic of a sample of atoms in theA. gas phaseB. liquid phaseC. solid phaseD. plasma state |
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Answer» Correct Answer - A Only atoms in the gas phase produce line spectrum. |
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| 529. |
Which of the following is correct? (i) A fluorescent lamp, like the ones commonly found in offices, is a discharged tube in which the inner surface is coated with a fluoresecnet material such as zinc sulphide which is also used on `TV` screen. (ii)The fluorescent tube is filled with ercury vapor at low pressure. Excitation of the `Hg` atoms by electron bombardment cause the emission of light in the green, blue, and ultraviolet `(UV)` regions. (iii) When the light strike the inner glass wall, most of the `UV` light is abosorbed by the fluorescent material, which then emits a multitude of longer wavelengths that combine to produce white light. (iv) Fluorescent lamps are more enegry-efficient and, hence, cheaper to operate than tungsten lamp (ordinary light bulbs).A. `(i),(iv)`B. `(ii),(iii),(iv)`C. `(i),(ii),(iii)`D. `(i),(ii),(iii),(iv)` |
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Answer» Correct Answer - D The generation of fluorescent light involves only the transfer of radiant enegry-very little heat is produced. As a result, a fluorscent lamp is cool to the touch. Heating the filament of a tungsten lamp to `3000^(@)C` produces visible light and much infrared radiation (heat). Thus, a tungsten light bulb is usually too hot to touch after it has been turned on for a few minutes because a substantial fraction of the input enegry is radiated as heat. |
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| 530. |
In which of the following pairs the ions are isoelectronic?A. `Na^(+),Mg^(2+)`B. `Al^(3+).O^(-)`C. `Na^(+),O^(2-)`D. `Na^(3-),Cl^(-)` |
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Answer» For `Na^(+)=11-1=10e^(-)` `Mg^(2+)=12-2=10e^(-)` Thus, they have same number of electorns. for, ` Al^(3+)=13-3=10e^(-),O^(-)=8+ 1=9e^(-)` They do not have same number of electrons For `Na^(+)=10e^(-),O^(2-)=8+2=10e^(-)` They have same number of electrons. For `N^(3-)=7+3=10e^(-).Cl^(-)=17+1=18e^(-)` They do not have same number of electrons Thus, `Na^(+)` is isolectronic with `Mg^(2+)` and `O^(2-)` |
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| 531. |
Which of the following sets of quantum numbers are correct? n l m n l mA. 1 1 +2B. 2 1 +1C. 3 2 -2D. 3 4 -2 |
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Answer» If `n=1,lcancel=(because lltn)` Hence, (a) is incorrect if `n=2,l=01, For l=1,m=-1.0.+1 Hence, (b) is correct If n=3,l=01.2 Hence (c) is correct if n=3, `lcancel=4` Hence (d) is incorrect. |
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| 532. |
Which of the following gases can be used to make neon signs? (i) Neon (ii) Argon (iii) KryptonA. `(i),(ii),(iii)`B. `(i)`C. `(i),(ii)`D. `(i),(iii)` |
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Answer» Correct Answer - A The familiaer neon signs we see on store fronts are colored by the atomic emission phenomenon. Excited neon atoms release orange-red light. Other gases may also be used to make 'neon' signs. For example, argon emits a bluish purple light and krypton gives off white light. special fluorescent materials on the inner glass walls of sign produce green and other colours as well. |
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| 533. |
The different orbits or energy levels of an atom according to Bohr model are called....... |
| Answer» Correct Answer - stationary states | |
| 534. |
Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c). radiation from FM radio (d) cosmic rays from outer space and (e) X-rays |
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Answer» The increasing order of frequency is as follows: Radiation from FM radio `l` amber light `lt` radiation from microwave oven `lt` X-rays `lt` cosmic rays The increasing order of wavelength is as follows: Comic rays `lt` X-rays `lt` radiation from microwave ovens `lt` amber light `lt` radiation of FM radio |
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| 535. |
Graph of incident freuency with stopping potential in photoelectric effect isA. B. C. D. |
| Answer» Correct Answer - 1 | |
| 536. |
`H_(α)` line of Balmer series is 6500 `Å` . The wave length of `H_(gamma)` isA. 4815 `Å`B. 4298`Å`C. `7800Å`D. `3800Å` |
| Answer» Correct Answer - 2 | |
| 537. |
Calculate the de Broglie wavelength of an electron moving at 1.5% of the speed of light. |
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Answer» Correct Answer - `1.62XX10^(-10)m` According to de Broglie equation , `lambda=(h)/(m v)` `m=9.1xx10^(-31) "kg", h=6.626xx10^(-34) "kg" m^(2)s^(-1)`, `v=(1.5xx3xx10^(8))/(100)=4.5xx10^(6) m s^(-1)` `lambda=((6.626xx10^(-34) "kg"m^(2)s^(-1)))/((9.1xx10^(-31) "kg")xx(4.5xx10^(6)ms^(-1)))=1.62xx10^(-10)m`. |
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| 538. |
What is the maximum number of emission lines obtained when the excited electron of a H atom in n = 5 drops to the ground state?A. 12B. 15C. 21D. 10 |
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Answer» Correct Answer - D Number of lines produced when electron from `n^(th)` shell drops to ground state is given by `(n(n-1))/2=(5(5-1))/2=10` |
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| 539. |
For principal quantum number, n = 4, the toal number of orbitals having l = 3 isA. 3B. 7C. 5D. 9 |
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Answer» Correct Answer - B For `l=3`(f-sub shell) No. of orbitals =7 |
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| 540. |
The velocity of an electron in a certain Bohr orbit of H-atom beaes the ratio 1 : 275 to the velocity of light. The quantum number (n) of the orbit isA. 3B. 2C. 1D. 4 |
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Answer» Correct Answer - B Velocity of electrons = `1/275xx " velocity of light " 1/275xx3xx10^(10)=1.09xx10^(8)cms^(-1)` Since `v_(n)=(2pie^(2))/(nh)` `therefore " " 1.09xx10^(8)=(2xx3.14xx(4.803xx10^(-10))^(2))/(6.626xx10^(-27)xxn)` ` therefore " "n=2` |
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| 541. |
The atomic spectrum of `Li^(2+)` arises due to the transition of an electron from `n_(2)` to `n_(1)` level. If `n_(1) +n_(2) ` is 4 and `n_(2)-n_(1)` is 2, calculate the wavelength (in nm) of the transition for this series in `Li^(2+)` |
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Answer» Step I. Calculation of value of `n_(1)` and `n_(2)` According to available information, `n_(1)+n_(2)=4` and ` n_(2)-n_(1)=2` On adding , ` 2n_(2)=6 " or " n_(2) =3` and ` n_(1)=4-3=1`. Step II. Calculation of wavelength of transition According to Rydberg formula, `(1)/(lambda)=bar(v)=R_(H)Z^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) , R_(H)=109678 cm^(-1) , Z(Li^(2+))=3, n_(1)=1, n_(2)=3` `:. " " (1)/(lambda) =(109678 cm^(-1))xx(3)^(2) ((1)/(1^(2))-(1)/3^(2))=(109678 cm^(-1)) xx 9 xx(8)/(9)= 877424 cm^(-1)` `lambda =(1)/((877424 cm^(-1)))=(1.14 xx 10^(-6)cm)= 1.14 xx 10^(-6) xx (1)/(10^(-7))=11.4 nm." " ` (`:. 1 nm=10^(-7)`cm) |
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| 542. |
Why does that splitting of spectral lines occur when the source of the spectrum is placed in a magnetic field ? |
| Answer» The splitting of the spectral lines known as Zeeman effect is an account of the reason that each sub shell has a number of orbitals. They acquire different orientations when the source giving the spectrum is placed under the influence of a strong magnetic field. These orientations correspond to different spectral lines. | |
| 543. |
In a certain electronic transition in Hydrogen atom from an initial state to a final state, the difference of orbit radius is 8 times the first Bohr radius. Which transition does not satisfy the given condition?A. `7 rarr 1`B. `6 rarr 1`C. ` 5 rarr1`D. `3 rarr1` |
| Answer» Correct Answer - A::B::C | |
| 544. |
The radius of `5^(@)` Bohr orbit in hydrogen atom isA. `5r_(0)`B. `(r_(0)) /( 5)`C. ` 25r_(0)`D. `(r_(0))/(25)` |
| Answer» Correct Answer - 3 | |
| 545. |
Calculate the wavelength of the spectral line obtained in the spectrum of `Li^(2+)` ion when the transition takes place between two levels whose sum is 4 and the difference is 2 |
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Answer» Suppose the transition takes place between levels `n_(1) and n_(2)` Then `n_(1) + n_(2) = 4 and n_(2) - n_(1) = 2` Solving these equations, we get `n_(1) = 1, n_(2) = 3 :. (1)/(lamda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) Z^(2)` For `Li^(2+) , Z = 3 :. (1)/(lamda) = 109, 677 cm^(-1) ((1)/(1^(2)) - (1)/(3^(2))) xx 3^(2) = 109, 677 xx ((1)/(1) - (1)/(9)) xx 9 cm^(-1) = 109677 xx 8 cm^(-1)` or `lamda = (1)/(109677 xx 8 cm^(-1)) = 1.14 xx 10^(-6) cm` |
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| 546. |
Calculate the wavelength of photon which will be emitted when the electron of hydrogen atom jumps from fourth shell to the first shell. Ionisation energy of hydrogen atom is `1.312xx10^(3)hJ "mol"^(-1)`. |
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Answer» Correct Answer - `9.725xx10^(-8)m` Step I. Amount of energy released lonisation energy of hydrogen atom `=1.312xx10^(3) kJ "mol"^(-1)` `=(1.312xx10^(3)xx10^(3))/(6.022xx10^(23))=2.18xx10^(-18) J "atom"^(-1)` Energy of electron in first shell `(E_(1))=E_(prop)-l.E.` `=0-2.18xx10^(-18) J "atm"^(-1)=-2.18xx10^(-18) J "atom"^(-1)` Energy of electron in fourth shell `(E_(4))=(-2.18xx10^(-18))/(n^(2))=(-2.18xx10^(-18))/((4)^(2))` `=0.136xx10^(-18) J "atom"^(-1)` Amount of energy released `(Delta E)=E_(4)-E_(1)=(-0.136xx10^(-18))-(-2.18xx10^(-18))` `=2.044xx10^(-18) J "atom"^(-1)` Step II. Wavelength of photon emitted Energy (E)=`hv=(hc)/(lambda) , lambda=(hc)/(E)=((6.626xx10^(-34)Js)(3xx10^(8)ms^(-1)))/((2.044xx10^(-18) J))=9.725xx10^(-8)m`. |
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| 547. |
The ionization energy of hydrogen in excited state is `+0.85 eV`. What will be the energy of the photon emitted when it returns to the ground state? |
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Answer» Energy of H-atom in the ground state `= - 13.6 eV` Ionization energy of hydrogen in excited state equal to `+0.85 eV` means Energy of H-atom in the excited state `= - 0.85 eV` `:.` Energy emitted `= -0.85 eV - (-1.6 eV) = 12.75 eV` |
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| 548. |
Calculate the number of photons emitted in `10` hours by a `60 W` sodium lamp (`lambda` of photon `= 5893 Å)` |
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Answer» Energy emitted by sodium lamp in 1 sec = `"watt" xx"sec"=60xx1=60 J`. Energy of photons emitted (E) = `hv=hc//lambda` `=((6.626xx10^(-34)Js)xx(3xx10^(8)ms^(-1)))/((5893xx10^(-10)m))=3.37xx10^(-19)J`. No. of photon emitted per second `=("Energy emitted in 1s")/("Energy of one photon")=((60J))/((3.37xx10^(-19)J))=1.78xx10^(20)`. No. of photons emitted in 10 hrs= `1.78xx10^(20)xx10xx60xx60=6.41xx10^(24)`. |
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| 549. |
Time taken by an electrons to complete one revolution in the Bohr orbit of the `H` atom isA. `(4pi^(2)mr^(2))/(nh)`B. `(nh)/(4pi^(2)mr)`C. `(2pimr)/(n^(2)h^(2))`D. `h/(2pimr)` |
| Answer» Correct Answer - A | |
| 550. |
Assertion. Hydrogen has one electron in its orbit but it produces several spectral lines Reason. There are many excited energy levels availableA. If both assertion and reason are true, and reason is the true explanation of the arrertionB. If both assertion and reason are true, but reason is not the true explanation of the assertionalC. If assertion is true, but reason is falseD. If both assertion and reason are false |
| Answer» Correct Answer - A | |