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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. |
The quantum number of electrons are given below: Arrange then in order of increasing energiesa.`n= 4,l= 2,m_(1)= -2, m_(s)= -(1)/(2)`b.`n= 3,l= 2,m_(1) = 1, m_(s)= +(1)/(2)` c.`n= 4,l= 1,m_(1) = 0, m_(s)= +(1)/(2)``e.n= 3,l= 2,m_(1)= -2, m_(s)= +(1)/(2)`f.`n= 4,l= 1,m_(1) = +1, m_(s)= +(1)/(2)` |
| Answer» `e lt b = d lt c = f lta` | |
| 402. |
When photon of energy `25eV` strike the surface of a metal A, the ejected photelectron have the maximum kinetic energy photoelectrons have the maximum kinetic energy `T_(A)eV` and de Brogle wavelength `lambda_(A)` .The another kinetic energy of photoelectrons liberated from another metal B by photons of energy `4.76 eV` is `T_(B) = (T_(A) = 1.50) eV` .If the de broglie wavelength of these photoelectrons is `lambda_(B) = 2 lambda_(A)` then i.` (W_(B))_(A) = 2.25 eV II. (W_(0))_(B) = 4.2 eV` III`T_(A) = 2.0 eV IV. T_(B) = 3.5 eV`A. The work function fo (A) is `2 . 25 eV`B. The work function of (B) is ` 3. 25 eV`C. ` T_A 2.00 eV`D. ` T_B = 2 . 75 eV` |
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Answer» Correct Answer - D ` u_A = h/(lambda_Am) :. U_(A)/u_(B) = ( lambda_A) = 2 ( :. lambda_B = 2 lambda_A)` ` u_B = h/(lambda_Bm)` Now ` T_(A)/T_(B) = mu_(A^(2)) /(mu_(B^(2))) = (4)/(1)` `T_(A) = T_(B)= 1. 50` `:. T_B =0. 50` ` T_A = 0. 50 + 1. 50 =.00 eV` Also ` 4. 25 = hv_(0_A) + T_A ,` ` : . hv_(0_A) = 4 . 25 - 2. 00 = 2. 25 eV` ` 4. 20 = hv_(0_B) + T_B ,` ` :. hv_(0_B) = 4 . 20 - 0. 50 = 3 70 eV`. |
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| 403. |
The correct order of deteasing energies mergies of the electrons is : A. `"Electron" 3 gt "Electron" 4 gt "Electron" 1 gt "Electron" 2`B. `"Electron" 4 gt "Electron" 3 gt "Electron" 1 gt "Electron" 2`C. `"Electron" 3 gt "Electron" 4 gt "Electron 2" gt "Electron" 1`D. `"Electron" 3 gt "Electron" 1gt "Electron" 4 gt "Electron "2` |
| Answer» `"Electron" 4 gt "Electron" 3 gtgt "Electron" 1 gt "Electron" 2` | |
| 404. |
Which of the following is(are) correct for if atom ? `1s lt 2s lt 2p lt 3s lt 3p` `1s lt 2s = 2p lt 3s = 3p` `1s lt 2p lt 3d lt 4s``1s lt 2s lt 4s lt 3d` The correct choice isA. ii,iiiB. I,ivC. I,iiiD. ii,iv |
| Answer» Correct Answer - I,iv | |
| 405. |
Ratio of wavelength of series limit of Paschen and Brackett series for a single electronic species is :A. `(4)/(9)`B. `(12)/(7)`C. `(9)/(16)`D. `(16)/(25)` |
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Answer» Correct Answer - C Correct option is C. \(\frac 9{16}\) As we know \(\frac 1\lambda = R_H \left(\frac 1{{n_f}^2}-\frac1{{n_i}^2}\right)\) .....(i) For limiting condition \(n_i= \infty\) for both Paschen and Brackett series \(n_f = 3\) for Paschen series \(n_f = 4\) for Brackett series \(\therefore \frac 1{\lambda_{Paschen}}= R_H\left(\frac1{3^2}- \frac1{\infty^2}\right)\) \(\frac 1{\lambda_{Paschen}}=\frac {R_H}9\) \(\because \left(\frac 1{\infty^2} \to0\right)\) \({\lambda_{Paschen}}=\frac 9{R_H}\) .......(ii) \(\therefore \frac 1{\lambda_{Brackett}}= R_H\left(\frac1{4^2}- \frac1{\infty^2}\right)\) \(\frac 1{\lambda_{Brackett}}= \frac{R_H}{16}\) \(\because \left(\frac 1{\infty^2} \to0\right)\) \({\lambda_{Brackett}}= \frac{16}{R_H}\) .......(iii) Therefore ratio of wave length of Paschen and Brackett series in limiting condition- \(\frac{\lambda_{Paschen}}{\lambda_{Brackett}} = \frac{9/R_H}{16/R_H} = \frac 9{16}\) |
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| 406. |
Excited atoms emit radiations consisting of only certain discrete frequencise or wavelengths. In spectroscopy it is often more convenient to use freuquencies and wave numbers are proportional ot energy and spectroscopy involves transitions between different energy levels . The line spectrum shown by a single electron excited atom (a finger print of an atom) can be given as `1/(lambda)=barvR_(H).Z^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` where Z is atomic number of single electron atom and `n_(1),n_(2)` are integers and if `n_(2)gt n_(1),` then emission spectrum is noticed and if `n_(2)ltn_(1),` then absorption spectrum is noticed. Every line in spectrum can be represented as a difference of two terms `(R_(H)Z^(2))/(n_(1)^(2)) "and"(R_(H)Z^(2))/(n_(2)^(2)).` H-atoms in ground state (13.6 eV) are excited by monochromatic radiations of photon of energy 12.09 eV. The number of different spectral lines emitted in H-atoms will be:A. 1B. 2C. 3D. 6 |
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Answer» Correct Answer - c |
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| 407. |
Excited atoms emit radiations consisting of only certain discrete frequencise or wavelengths. In spectroscopy it is often more convenient to use freuquencies and wave numbers are proportional ot energy and spectroscopy involves transitions between different energy levels . The line spectrum shown by a single electron excited atom (a finger print of an atom) can be given as `1/(lambda)=barvR_(H).Z^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` where Z is atomic number of single electron atom and `n_(1),n_(2)` are integers and if `n_(2)gt n_(1),` then emission spectrum is noticed and if `n_(2)ltn_(1),` then absorption spectrum is noticed. Every line in spectrum can be represented as a difference of two terms `(R_(H)Z^(2))/(n_(1)^(2)) "and"(R_(H)Z^(2))/(n_(2)^(2)).` The ratio of wavelength for II line fo Balmer series and I line of Lyman series for H-like species is :A. `1/4`B. 2C. 3D. 4 |
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Answer» Correct Answer - d |
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| 408. |
Which of the following transitions of an electron in hydrogen atom emits radiation of the lowest wavelength?A. `n_(2)=infty` to `n_(1)=2`B. `n_(2)=4` to `n_(1)=3`C. `n_(2)=2` to `n_(1)=1`D. `n_(2)=5` to `n_(1)=3` |
| Answer» Correct Answer - A | |
| 409. |
If `lamda_o and lamda` be the threshold wavelength and wavelength of incident light , the velocity of photoelectron ejected from the metal surface is :A. `sqrt((2h)/(m)(lamda_(0)-lamda)`B. `sqrt((2hc)/(m)(lamda_(0)-lamda))`C. `sqrt((2hc)/(m)((lamda_(0)-lamda)/(lamdalamda_(0))))`D. `sqrt((2h)/(m)((1)/(lamda_(0))-(1)/(lamda)))` |
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Answer» Correct Answer - C Absorbed energy=Threshold energy+kinetic energy of photoelectrons `(hc)/(lamda)=(hc)/(lamda_(0))+(1)/(2)mv^(2)` `v=sqrt((2hc)/(m)((lamda_(0)-lamda))/(lamdalamda_(0)))`. |
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| 410. |
The work function of a metal is `4.0 eV`. If the metal is irradiated with radiation of wavelength `200` nm, then the maximum kinetic energy of the photoelectrons would be about.A. `6.4 xx 10^-19 J`B. `3.5 xx 10^-19 J`C. `1.0 xx 10^-18 J`D. `2.0 xx 10^-19 J` |
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Answer» Correct Answer - B (b) `W_0 = 4.0 eV = (hc)/(lamda_0)` =`4 xx 1.6 xx 10^-19 J` `n = 3 xx 10^15 Hz` `6.4 xx 10^-19 J` Now apply `KE = (hc)/(lamda) - (hc)/(lamda_0)` when given `lamda = 200 nm = 2 xx 10^-7 m`. |
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| 411. |
A metal is irradiated with light of wavelength 660 nm. Given that the work that the work function of the metal is `1.0eV,` the de Broglie wavelength of the ejected electron is close to-A. `6.6xx10^(-7)m`B. `8.9xx10^(-11)m`C. `1.3xx10^(-9)m`D. `6.6xx10^(-13)m` |
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Answer» Correct Answer - C `E_("Absorbed")=(hc)/(lamda)=(12400)/(600)=1.88eV` `E_("Absorbed")=E_(0)+KE` `KE=E_("absorbed")-E_(0)` `=(1.88-1)eV=0.88eV` de Broglie wavelength`lamda=(h)/(sqrt(2Em))=sqrt((150)/(KE(eV)))` `=sqrt((150)/(0.88))=sqrt(170.45)` `=13"Ã…"=1.3xx10^(-9)m` |
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| 412. |
Threshold wavelength of a metal is `lamda_(0)`. The de Broglie wavelength of photoelectron when the metal is irradiated with the radiation of wavelength `lamda` is:A. `[(h lamdalamda_(0))/(2cm)]^((1)/(2))`B. `[(h(lamda-lamda_(0)))/(2cmlamdalamda_(0))]^(1//2)`C. `[(h(lamda_(0)-lamda))/(2cmlamdalamda_(0))]^(1//2)`D. `[(hlamdalamda_(0))/(2mc(lamda_(0)-lamda))]^(1//2)` |
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Answer» Correct Answer - D Absorbed energy=Threshold energy+Kinetic energy of photoelectron `h(c)/(lamda)=(hc)/(lamda_(0))+E` `E=hc[(1)/(lamda)-(1)/(lamda_(0))]` . . . .(i) de Broglie wavelength can be calculated as `lamda=(h)/(sqrt(2Em))` .. . . (ii) From eqs. (i) and (ii) `lamda=[(hlamdalamda_(0))/(2mc(lamda_(0)-lamda))]^(1//2)` |
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| 413. |
When beryllium is bombarded with `alpha-"particles"`, extremely penetrating radiations which cannot be deflected by electrical or magnetic field are given out. These areA. A beam of protonsB. `alpha` - raysC. A beam of neutronsD. X-rays |
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Answer» Correct Answer - C Neutron is a chargeless particles, so it does not get deflected by electric or magnetic field. |
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| 414. |
When atoms are bombarded with alpha particles, only a few in million suffer deflection, others pass out undeflected. This is becauseA. The force of repulsion on the moving alpha particle is smallB. The force of attraction on the alpha particle to the oppositely charged electrons is very smallC. There is only one nucleus and large number of electronsD. The nucleus occupies much smaller volume compared to the volume of the atom |
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Answer» Correct Answer - D The nucleus occupies much smaller volume compared to the volume of the atom |
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| 415. |
When atoms are bombarded with alpha particles, only a few in millon suffer deflection, other pass out undeflected. This is because.A. The force of repulsion on the moving alpha particle is smallB. The force of attraction on the alpha particle to the oppositely charged electrons is very smallC. There is only one nucleus and large number of electronsD. The nucleus occupies much smaller volume compared to the volume of the atom. |
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Answer» Correct Answer - D (d) The nucleus occupies much smaller volume compared to the volume of the atom. |
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| 416. |
Show that the wavelength of a moving particle is relased to its kinetic energy `(E ) as lambda = (h)/(2mE)^(1//2)` |
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Answer» Kinetic energy `E= (1)/(2) mv^(2)` `2mE =- (1)/(2) mv^(2) xx 2m = m^(2)v^(2)` `or mv = (2 mE)^(1//2)` According to de Broglie, `lambda = (h)/(mv) = (h)/((2mE)^((1)/(2)))` |
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| 417. |
Positronium consists of an electron and a positron (a particle which has the same mass as an electron , but opposite charge ) orbiting round their common centre of mass. Calculate the value of the Rydberg constant for this systemA. `R_oo //4`B. `R_oo //2`C. `2R_oo`D. `R_oo` |
| Answer» Correct Answer - B | |
| 418. |
What is the ratio of the velocities of `CH_(4)` and `O_(2)` molecules no that they are associated with de broglie waves of equal wavelength ? |
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Answer» From the de Brogle relationship `lambda = (h)/(mv)` For `CH_(4) ,lambda_(ch4) = (h)/(m_(CH4) xx v_(CH4))`…….(i) For `O_(2),lambda_(O_2) = (h)/(m_(O_2) xx v_(O_2))`……(ii) Wavelength of `CH_(4)` and `O_(2)` is equal hence ` (h)/(m_(CH_4) xx v_(CH_4)) =(h)/(m_(O_2) xx v_(O_2))` `rArr ( v_(CH_4))/( v_(O_2))= (m_(O_2))/(mCH_4) = (32)/(16) = 2 ` `:, v_(CH_4) = 2 v_(O_2)` THe velocity of `CH_(4)` molecule is two times the velocity of `O_(2)`molecule |
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| 419. |
For`He^(Theta)` and `Li^(2+)` , the energies are relased to the quantum number n through an expression `E_(n) = (Z^(2)B)/(n^(2))` where Z is the atomic number species and `B = 2.179 xx 10^(-19) J` a.What is the energy of the lowest level of a `He^(Theta)` ion ? b. .What is the energy of the third level of a `Li^(2+)` ion ? |
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Answer» `E_(1)` for `He^(o+) = - (2^(2) xx2.179 xx 10^(-18))/(1^(2))` `= - 8.716 xx 10^(-18) J` b. `E_(3)` = for `LI^(2+) = - (3^(2)xx2.179 xx 10^(-18))/(3^(2))` `= - 2.179 xx 10^(-18) J` |
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| 420. |
What should be the ratio of the velocity of `1CH_(4) and O_(2)` molecules so that they are associated with de Broglie wave of equal wavelegth ? |
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Answer» According to de Broglie equation For methane `(CH_(4))` `lambda _(CH_4) = (h)/(m_(CH_4) xx v_(CH_4))` For oxygen `(O_(2))` `lambda_(O_2) = (h)/(m_(O_2) xx v_(O_2))` since the wavelength of `CH_(4)` and `O_(2)` sould be equal `lambda _(CH_4) = lambda _(O_2)` `or (h)/(m_(CH_4) xx v_(CH_4)) = (h)/(m_(O_2) xx vO_(2))` `m_(CH_4) xx vCH_(4) = m_(O_2) xx vO_(2)` `or (vCH_4)/(v_(O_2)) = (m_(O_2))/(m_(CH_4)) = (32)/(16) = 2` |
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| 421. |
The mathematical expression for the uncertainty principle isA. `Delta x Delta p ge (h)/(4pi)`B. `Delta E Delta r ge (h)/(4pi)`C. `Delta x Delta p ge (h)/(p)`D. `Delta E Delta r ge (h)/(p)` |
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Answer» `Delta x Delta p ge (h)/(4pi) b. We know that `Delta E = m Delta I^(2) = m((Delta x)/(Delta t))^(2)` `or Delta E -Delta t = m(Delta x^(2))/(Delta t)` `Delta x Delta p gt (h)/(4pi)` or `Delta x m Delta V ge (h)/(4pi)` `rArr Delta x m (Delta x)/(Delta t) ge (h)/(4pi)` ` rArr m (Delta x)/(Delta t) ge (h)/(4pi)` On compairing (i) and (ii) we get `delta E = Delta t ge (h)/(4pi)` |
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| 422. |
Calculate the energy emitted when electrons of `1.0 g`1 of hydrogen transition giving spectrum lines of the lowest in the visible regain of its atomic spectrum `R_(H) = 1.1 xx 10^(7) m^(-1) , c= 3 xx 10^(8) m s^(-1)` and `h = 6.62 xx 10^(-34) J s` |
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Answer» For visible line spectrum, i.e. Balmer series `n_(1) = 2` Also for minimum energy transion `n_(2) = 3` so `:. (1)/(lambda) = R_(H)((1)/(n_(1)^(2))- (1)/(n_(2)^(2)))` `therefore1/lambda=R_(H)(1/2^(2)-1/3^(2))` `= 1.1 xx 10^(7)((1)/(4)-(1)/(9)) = 1.1xx 10^(7) xx (5)/(36)m^(-1)` `:. lambda = 6.55 xx 10^(-7) m` `E = (hc)/(lambda)= (6.62 xx 3.0 xx 10^(8))/(6.55 xx 10^(7)) = 3.037 xx 10^(-19) J` If N elecytron show the transition in `1 g of H` atom , then energy released `= E xx N` `= 3.03 xx 10^(-19) xx 6.023 xx 10^(23)` `= 18.29 xx 10^(6) J = 182.5 J` |
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| 423. |
Calculate the energy emitted when electrons of `1.0 g` of hydrogen undergo transition giving spectrum lines of the lowest energy in the visible region of its atomic spectrum. `R_(H) = 1.1 xx 10^(7) m^(-1) , c= 3 xx 10^(8) m s^(-1)` and `h = 6.62 xx 10^(-34) J s` |
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Answer» For visible line spectrum , i.e. Boalumer series ` n_1 =2`1 Also for minimum energy transition ` n_2 =3` . ` :. 1/( lambda) = R_H [ 1/n_1^2 - 1/n_2^2]` ` :. 1/(lambda) = RH [ 1/2^2 - 1/3^2 ]` ` -= 1.1 xx 10^7 [ 1/4 - 1/9] = 1.1 xx 10^7 xx 5/( 36)` ` :. lambda = 6. 55 xx 10^(-7) ` meter Now ` E= (hc)/( lambda) = ( 6.63 xx 10 10^(-34) xx 3 xx 10^8)/( 6. 55 xx 10^(-70)` ` = 3 . 03 xx 10 ^(-19) ` joule If (N) electrons shown this transition in 1 g-atom of (H) then Energy released ` = E xx N` ` = 3. 03 xx 10 ^(-19) xx 6.023 xx 10^(23)` ` = 18 . 25 xx 10^4 J = 182 . 5 k J`. |
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| 424. |
Predict the total spin in ` Ni^(2+) ` ion :A. `+-(5)/(2)`B. `+-(3)/(2)`C. `+-(1)/(2)`D. `+-1` |
| Answer» Correct Answer - 4 | |
| 425. |
In which of the following transition, the wavelength will be minimum `:`A. n=6 to n=4B. n=4 to n=2C. n=3 ton=1D. n=2 to n=1 |
| Answer» Correct Answer - 3 | |
| 426. |
Hydrogen atom electron is excited to n = 4 state , in the spectrum of emitted radiation , number of lines in the ultravoilet and visible regions are respectively :-A. 3 , 1B. 1 , 3C. 2 , 3D. 3 , 2 |
| Answer» Correct Answer - 4 | |
| 427. |
The first emission line of Balmer series in `He^(+)` - spectrum has the wave no .in `(Cm^(_1))` (R-Rydberg constant)A. `(3R)/(4)`B. `(20R)/(36)`C. `(5R)/(36)`D. `(R)/(6)` |
| Answer» Correct Answer - 2 | |
| 428. |
The first emission line in the H-atom spectrum in the Balmer series appears at:A. `(5R)/(36)cm^(-1)`B. `(3R)/(4)cm^(-1)`C. `(7R)/(144)cm^(-1)`D. `(9R)/(400)m^(-1)` |
| Answer» Correct Answer - A | |
| 429. |
The first emission line of Balmer series in H spectrum has the wave number equal to :A. `(9R_H)/(400) cm^(-1)`B. `(9R_H)/(144)cm^(-1)`C. `(3R_H)/4cm^(-1)`D. `(5R_H)/(36) cm^(-1)` |
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Answer» Correct Answer - D `overlinev = 1/(lambda)= R_H [ 1/2^2 - 1/3^2], n_1 = 2` for Balmer series , ` n_2 = 3` for `H_ (alpha)` line . |
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| 430. |
Which sub-shell has maximum energy (poly-elecronic system ) :A. `4f`B. `6s`C. `5d`D. `5p` |
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Answer» Correct Answer - C ` n+l` value for `5d` and ` 4f` is ` 7.` if `b+l` is same higher value of `n` is same higher value of n shows higher energy . |
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| 431. |
Yellow light emitted from a sodium lamp has a wavelength ( `lambda` ) of ` 580 nm`. Calculate the frequency (v). Wave number and energy of yellow light photon . |
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Answer» Frequency `(v) = c/( lambda) = ( 3. 0 xx 10^8)/( 580 xx 10^(-9)` ` = 5 . 17 xx 10^(14) "Hz or" s^(-1)` Wave no. `overline(v) = 1/(lambda) = 1/( 580 xx 120^(-9)) = 1.724 xx 10^6 m^(-1)` `E = hv = 6.625 xx 10^(-32) xx 5. 17 xx 10^(14)` ` 34 . 24 xx 10^(-20) J//phot on`. |
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| 432. |
Gases begin to conduct electricity at low pressure becauseA. at low pressures gases turn to plasmaB. colliding electrons can acquire higher kinetic energy due to increased mean free path leading to ionisation of atomsC. atoms break up into electrons and protonsD. the electrons in atoms can move freely at low pressure. |
| Answer» Correct Answer - B | |
| 433. |
Calculate frequency energy and wavelength of the radiation corresponding to the speciral line of the lowest frequency in lyman series in the spectrum of a hydrogen atom .Also calculate the energy for the coresponding line in the spectrum of `Li^(2+).(R_(H) = 109.677 cm^(-1), c = 3 xx 10^(6) m s^(-1), Z = 3)` |
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Answer» In Lyman series `n_(1) = 1 and n_(2) = 2` (coreresponding lowest frequency region ) `(1)/(lambda) = bar v = R [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))] = 1.09677 xx 10^(7) m^(-1)[(1)/(1^(2))- (1)/(2^(2))]` `= 1.09677 xx 10^(7) m^(-1)(1 - (1)/(4)) = (1.9677 xx 10^ (7) xx 3)/(4) cm ` `:. lambda = (4)/(1.09677 xx 10^(7) xx 3 ) = 1.215 xx 10^(7) m` `"Frequency" v = (c )/(lambda) = (3 xx 10^(8) ms^(-1))/(1.215 xx 10^(-7) m) = 2.4677 xx 10^(15)s^(-1)` Energy of radiation `E = hv` `:. E = 6.625 xx 10^(-34) Js xx 2.4677 xx 10^(15)s^(-1)` `= 1.6348 xx 10^(-16) J` energy for corresponding line in the spectral of `Li^(2+)` = Energy for hydrogen atom `xx Z^(2) = 1.6348 xx 10^(-18)J xx 3^(2) = 1.471 xx 10^(-17)J` |
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| 434. |
Calculate frequency, energy and wavelength of the radiation corresponding to the speciral line of the lowest frequency in lyman series in the spectrum of a hydrogen atom .Also calculate the energy for the coresponding line in the spectrum of `Li^(2+).(R_(H) = 109677 cm^(-1), c = 3 xx 10^(8) m s^(-1), Z = 3)` |
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Answer» Correct Answer - A::B::C::D `(1)/(lambda) = bar v = R [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` `= 1.09678 xx 10^(7)m^(-1)[(1)/(1^(2)) -(1)/(2^(2))]` `= 1.09678 xx 10^(7)m^(-1)(1 - (1)/(4))` `lambda = (4)/(1.09678 xx 10^(7) xx 3)m^(-1) = 1.2157 xx 10^(-7)m` ` c/lambda=(3xx10^(8)ms^(-1))/(1.2157xx10^(-7)m)=2.4677xx10^(15)s^(-1)` `E = hv = 6.6625 xx 10^(-14) Js xx 2.4677 xx 10^(15) s^(-1) = 1.63485 xx 10^(18) J` [ `E = hv = 6.6625 xx 10^(-14) Js xx 2.4677 xx 10^(15) s^(-1) = 1.63485 xx 10^(18) J` [Energy for comes- ponding line spetral of `Li^(2+)]` = [Energy for hydrogen atom `xx Z^(2)]` `1.63485 xx 10^(-10) J xx 3^(2)` ` = 1.471.365 xx 10^(-17) J` |
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| 435. |
The angular momentum (L) of an electron in a Bohr orbit is gives as:A. `L=(nh)/(2pi)`B. `L=sqrt(l(l+1)(h)/(2pi))`C. `L=(mg)/(2pi)`D. `L=(h)/(4pi)` |
| Answer» Correct Answer - A | |
| 436. |
To what series does the speciral lines of atomic hydrogen belong if its wavelength is equal to the difference between the wavenumber of the folowing two lines of the Balmer series `486.1` and `419.2 nm`? What is the waveeath of thqat line ? |
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Answer» Given `lambda_(1) = 486.1 xx 10^(-9)m = 486.1 xx 10^(-7) cm` `lambda_(2) = 410.2 xx 10^(-9)m = 410.2 xx 10^(-7) cm` ` bar v = bar v_(2) - barv _(1)` `= (1)/(lambda_(2)) - (1)/(lambda_(1))` `= R_(H) [(1)/(2^(2)) - (1)/(n_(2)^(2))] - R_(H) [(1)/(2^(2)) - (1)/(n_(1)^(2))]` `bar v = R_(H)[1/n_(1)^(2)-1/n_(1)^(2)]`........(i) For 1 case of balance series `(1)/(lambda_(1)) = R_(H) [(1)/(2^(2)) - (1)/(n_(1)^(2))] = 109678 [(1)/(2^(2)) - (1)/(n_(1)^(2))]` or `(1)/(4186.1 xx 10^(-5)) = 109678[(1)/(2^(2)) - (1)/(n_(1)^(2))]` `:. n_(1) = 4` For II case of Balmer series `(1)/(lambda) = (1)/(410.2 xx 10^(-7)) = 10978 [(1)/(2^(2)) - (1)/(n_(2)^(2))]` `:. n_(2) = 6` THus giuven transition occurs from sixth level to four level .Also equation (i) `bar v = (1)/(lambda) = 109678 [(1)/(4^(2)) - (1)/(6^(2))]` `:. lambda = 2.63 xx 10^(-4) cm ` |
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| 437. |
Calculate (a) Wavenumber and (b) frequency of yellow radiation having wavelength `5800 A^(@)`. |
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Answer» (a) Calculate of wavenumber `(bar(v))` `lambda = 5800 A^(@) = 5800 xx 10^(-8) cm = 5800 xx 10^(-10)m` `bar(v) = (1)/(lambda) = (1)/(5800 xx 10^(-10)m) = 1.724 xx 10^(6)m^(-1)` `= 1.724 xx 10^(4)cm^(-1)` (b) Calculation of the frequency (V) `v = (c)/(lambda) = (3 xx 10^(8)ms^(-1))/(5800 xx 10^(-10)m) = 5.172 xx 10^(14)s^(-1)` |
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| 438. |
a. An atomic orbit has `n= 3` What are the possible values of i? b. An atomic orbital has `l = 3` when are the possible value of m ? |
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Answer» a. When `n = 3,l = 0,1,2` b. When `l = 3,m= p-3,m = -3,-2 -1,0,1,2,3,` |
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| 439. |
The z-component of angular momentum of an electron in an atomic orbit is government by theA. Principal quantum numberB. Azamothal quantum numberC. Magnetic quantum numberD. Spin quantum number |
| Answer» Magnetic quantum number | |
| 440. |
Calculate the apperomixmate of polonium `210` nucless |
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Answer» Correct Answer - A::C::D Radiation of nucless can be esimated by `R = 1.4 xx 10^(-13) (A)^((1)/(3)) `(A is the mass number ) `= 1.31 xx 10^(-13) (210)^(1//3) = 8.3 xx 10^(-15) cm` |
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| 441. |
Find the wavenumber curresponding to the longest wavelength photon to remove electron from the second excited state of `He^(o+) "ion" (R= 1.097 xx 10^(7)m^(-1))` |
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Answer» Second excited state `-=n = 3` Longest wavelength photon refers to transition `n = 3 to 4` `bar v = RZ^(2)[(1)/(3^(2)) - (1)/(4^(2))]` `= 1.097 xx 10^(7)(4)[(1)/(9) - (1)/(16)]m^(-1)` `= 2.133 xx 10^(6)m^(-1)` |
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| 442. |
The energy required to ionize a potassium ion is 419 kJ `mol^(-1)` . What is the longest wavelength of light that can cause this ionization ?A. 285 nmB. 216 nmC. 200 nmD. 107 nm |
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Answer» Correct Answer - A |
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| 443. |
How many chlorine atoms can you ionize in the process `Cl rarr Cl^+ + e`, by the energy liberated from the following process ? `Cl + e^- rarr Cl^(-) for 6 xx 10^23 "atoms"` Given electron affinity ofm `Cl = 3.61 eV`, and `I P` of `Cl = 17.422 eV`.A. `1.24 xx 10^23 "atoms"`B. `9.82 xx 10^20 "atoms"`C. `2.02 xx 10^15 "atoms"`D. none of these |
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Answer» Correct Answer - A (a) Energy released in conversion of `6 xx 10^23 "atoms"` os `Cl^(-) ions` =`6 xx 10^23 xx "electron affinity"`, =`6 xx 10^23 xx 3.61` =`2.166 xx 10^24 eV` Let `x Cl` atoms are converted to `Cl^+` ion Energy absorbed `= x xx "ionization energy"` `x xx 17.422 = 2.166 xx 10^24` `x = 1.243 xx 10^23 "atoms"`. |
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| 444. |
The de Broglie wavelength of an electron moving in a circular orbit is `lambda` . The minimum radius of orbit is :A. `(lambda)/(4)`B. `(lambda)/(2pi)`C. `(lambda)/(4 pi)`D. `(lambda)/(3 pi)` |
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Answer» Correct Answer - B |
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| 445. |
A particle of `(e)/(m)` ratio equal to `4 xx 10^(5)` coul/kg is accelerated from rest through a potential difference of 20 volt . The speed of particle is :A. 4.0 m/sB. 4000 km/sC. 4000 cm/sD. 4000 m/s |
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Answer» Correct Answer - D |
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| 446. |
The radial probability is the probability of finding electron in a small spherical shell around the nucless at a particular distance ® Hence radial probility isA. `4pi r^(2) dr Psi^(2)`B. `(4//3) pi r^(2) dr Psi^(2)`C. `2pi r^(2) dr Psi^(2)`D. `4pi r dr Psi` |
| Answer» `R = 4pi r^(2) dr v^(2)` | |
| 447. |
A single electron orbits around a stationary nucless of charge `+ Ze` where `Z` is a constant and e is the magnitude of electronic charge ,if respuires 47.2 e is excite the electron from the second bohr orbit to the third bohr orbit a. Find the value of Z b. Find the energy required to the electron from `n = 3" to" n = 4` c. Find the wavelength of radiation to remove the electron from the second bohr orbit to infinity d. Find the kinetic energy potential energy and angular momentum of the electron in the first orbit Find the ionination energy of above electron system in electronvalt. |
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Answer» Since the nucless has a chaarge `+ Ze`, the atomic nucless of the ion is Z a. The transition is `n_(!) = 2 rarr n_(2) = 3 ` by abserbing a photon of energy `47.2 eV` `rArr Delta E = 47.2 eV` Using the relation : `Delta E = 13.6Z^(2)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))eV` `rArr 47.2 = 13.6Z^(2)((1)/(Z^(2)) - (1)/(3^2)) rArr Z = 3` The required transition is `n_(1) = 3 rarr n_(2) = 4` by abserbing aphoton of energy of nergy `Delta E` Find `Delta E`by using the relation `Delta E=13.6Z^(2)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` `rArr Delta E=13..6(5)^(2)((1)/(3^(2)) - (1)/(4^(2)))` `Delta E = 16.53 eV` c. The required transition is `n_(1) = 2 rarr n_(2) rarr n_(2) = 4`by abserbed aphoton of energy` Delata E ` Find `Delta E`by using the relation `Delta E=13..6(5)^(2)((1)/(Z^(2))- (1)/(oo^(2)))` `rArr Delta E = 8.5 eV` Find the `lambda` of relation corresponding to energy `85eV` `rArr lambda =- (hc)/(E ) = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(85(1.6 xx 10^(-19)))` `= 146.25 xx 10^(-10) = 146.25 Å` d. If energy of electron be `E_(n)` then `KE = - E_(n) ` and `PE = 2E_(n)` `E_(n) = (-13.6Z^(2))/(n^(2)) = (-13.65 xx 5^(2))/(1^(2)) = - 340 eV` `KE = (-340 ev) = 340 ev` `PE = 2(-34 eV) = -680 eV` e. Ionization energy `= Delta E = 13.6(5)^(2)((1)/(2^(2)) - (1)/(oo^(2)))` `:. Delta E = 85 eV` |
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| 448. |
`._(4)Be^(7)` capaires a K electron into its nucless .What is the mass number and atomic number of the nuclide formed ? |
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Answer» `_(4)Be^(7) + _(-1)e^(0) rarr _(5)LI^(7)` Atomic number `= 3`mass number `= 7` |
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| 449. |
Suppose a gaseous mixtures of `He, Ne, Ar ` and `Kr` is irradiated with photons for frequency appropriate to ionize Ar, What ions will be present in the mixture ?A. `He^(+)` onlyB. `Ne^(+)` onlyC. `He^(+) , Ne^(+) , Ar^(+)` onlyD. `Ne^(+) , Kr^(+) , Ar^(+) ` only |
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Answer» Correct Answer - D |
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| 450. |
The first ionization energy of caesium is `6.24 xx 10^(-19)J//"atom"` . What is the minimum frequency of light that is required to ionize a caesium atom ?A. `1.06 xx 10^(-15)s^(-1)`B. `4.13 xx 10^(14) s^(-1)`C. `9.42 xx 10^(14) s^(-1)`D. `1.60 xx 10^(18) s^(-1)` |
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Answer» Correct Answer - C |
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