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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.
| 101. |
Name the quantum number which does not follow from the solution of Schrodinger wave equation ? |
| Answer» Spin quantum no. `(m_(s))` does not follow from the Schrodinger wave equantion. | |
| 102. |
Calculate and compare the energies of two radiations one with wavelength 800 pm and the other with wavelength 400 pm. |
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Answer» `E=hv=(hc)/(lambda)` `"First case": " "lambda=800"pm"=800xx10^(-12)=8xx10^(-10)m` `c=3.0xx10^(8)m s^(-1), h=6.626xx10^(-34)Js` `:. " "E_(1)=((6.626xx10^(-34)Js)xx(3xx10^(8)ms^(-1)))/((4xx10^(-10)m))=4.96xx10^(-16)J` `"Second case": " "lambda=400"pm"=400xx10^(-12)=4xx10^(-10)m` `" "c=3.0xx10^(8)m s^(-1), h=6.626xx10^(-34)Js` `:. " "E_(2)=((6.626xx10^(-34)Js)xx(3xx10^(8)ms^(-1)))/((4xx10^(-10)m))=4.96xx10^(-16)J` `" " (E_(1))/(E_(2))=(2.48xx10^(-16)J)/(4.96xx10^(-16)J)=(1)/(2) or E_(2)=2E_(1)`. |
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| 103. |
Calculate the frequency and energy of a photon of radiation having wavelength `6000 Å` |
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Answer» Frequency `v = (c)/(lamda)` Substituting `c = 3 xx 10^(8) m s^(-1), lamda = 6000 Å= 6000 xx 10^(-10) m`, we get `v = (3 xx 10^(8) ms^(-1))/(6000 xx 10^(-10)m) = 5 xx 10^(4) s^(-1)` (ii) Energy of the photon E = hv Substituting `h = 6.625 xx 10^(-34) J s, v = 5 xx 10^(14) s^(-1)`, we get `E = 6.625 xx 10^(-34) J s xx 5 xx 10^(14) s^(-1) = 3.3125 xx 10^(-19) J` |
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| 104. |
The threshold energy for photoelectric emission of electrons from a metal is `3.056 xx 10^(-15)` joule. If light of `4000 Å` wavelength is used, will the electrons be efected or not ? (`h = 6.63 xx 10^(-34)` Joule sec) |
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Answer» Correct Answer - No Electrons will be ejected only if the energy of incident light is greater than the threshold energy. Energy of the incident light `= hv = h (c)/(lamda) = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(4000 xx 10^(-10)) J = 4.97 xx 10^(-19) J` Thus is less than threshold energy |
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| 105. |
The energy difference between ground state of an atom and its excited state is `3xx10^(-19)J`. What is the wavelength of photom required for this radiation ? |
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Answer» Correct Answer - `6.6xx10^(-7)m` `DeltaE = (hc)/(lambda)` `lambda=(hc)/(DeltaE)=((6.626xx10^(-34)Js)(3xx10^(8)ms^(-1)))/((3xx10^(19)J))=6.6xx10^(-7) m` |
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| 106. |
What is the number of photons of light with wavelength 4000 pm which provide 1 Joule of energy ? |
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Answer» Energy of photon (E) `=(hc)/(lambda)` `h=6.626 xx 10^(-34)Js, c=310^(8) ms^(-1), lambda=4000 xx10^(-12) =4xx10^(-9)m` `:. " " "Energy of photon" (E)=((6.626 xx10^(-34)Js)xx(3xx10^(8)m s^(-1)))/((4xx10^(-9)m))=4.969 xx 10^(-17)J` Now, `4.965 xx 10^(-17)J` is the energy of photon =1 `:. 1 J ` is energy of photons `=(1)/(4.969 xx10^(-17))=2.012 xx 10^(16)` photons. |
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| 107. |
what is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy ? |
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Answer» Energy (E) of a photon = hν Energy `(E_(n))` of ‘n’ photons = nhv `Rightarrow n=( E_(n)lamda)/(hc)` Where `lambda` = wavelength of light = 4000 pm = `4000xx10^(12) m` c = velocity of light in vacuum = ` 3xx 10^(8)m//s` h = Planck’s constant = `6.626 xx 10 ^(-34)Js` Substituting the values in the given expression of n: `n= ((1)xx(4000xx10^(-12)))/((6.626xx10^(-34)(3 xx 10^(8)))=2.012xx10^(16)` Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are `2.012 xx 10^(16)` |
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| 108. |
Which of the following sets of quantum numbers are correct ?A. `{:(n,l,m_(1)),(1,1,+2):}`B. `{:(n,l,m_(1)),(2,1,+1):}`C. `{:(n,l,m_(1)),(3,2,-2):}`D. `{:(n,l,m_(1)),(3,4,-2):}` |
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Answer» Correct Answer - B::C If `n=1, l != 1`. Hence (a) is wrong If `n=2, l =0, 1`. For `l = 1, m = -1, 0 +1`. Hence (b) is correct If `n=3, l =0, 1, 2`. For `l =2, m=-2, -1, =, +1, 2`. Hence (c) is correct If `n=3, l != 4`. Hence, (d) is wrong |
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| 109. |
What are thhe atomic numbers of elements whose outermost electrons are represented by (i) `3s^(1)` (ii) `2p^(3)` and (iii) `3d^(6)` ? |
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Answer» Correct Answer - (i) 11 (ii) 7 (iii) 26 (i) When last shell is `3s^(1)`, complete E.C.`=1s^(2)2s^(2) 2p^(6) 3s^(1)` (ii) When last shell is `2p^(3)`, complete E.C. `=1s^(2) 2s^(2) 2p^(3)` (iii) When last shell is `3d^(6)`, complete E.C.`=1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 4s^(2) 3d^(6)`. |
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| 110. |
The increasing order (lowest first) for the values of `e//m` (charge//mass) for electron `(e)`, proton `(p)`, neutron `(n)`, and alpha particle `(alpha)` isA. e, p, n, `alpha`B. n, p, e, `alpha`C. n, p, `alpha`, eD. n, `alpha`, p, e |
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Answer» Correct Answer - D e/m value for neutron (0/1)=0 `alpha`-particle (2/4)=0.5 proton (1/1)=1 electron `((1)/(1//1837))=1837`. ltbtgt `:.` (d) represents the correct increasing order. |
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| 111. |
The increasing order (lowest first) for the values of `e//m` (charge//mass) for electron `(e)`, proton `(p)`, neutron `(n)`, and alpha particle `(alpha)` isA. `n, p, alpha ,e`B. `e, p, n, alpha`C. `n, p, e, alpha`D. `n, alpha, p, e` |
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Answer» Correct Answer - D `e//m` for electron is `1//0` (as electron has negligible mass), for proton is `1//1`, for neutron is `0//1`, and for alpha particle is `2//4` or `1//2`. Note that charge unit for `e` is `-1`, for proton is `+1`, for neutron is `0`, and for alpha particle is `+2`. The mass unit (amu) for `e` is almost `0`, fro `p` is `1`, for neutron is `1`, and for alpha particles is `4`. |
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| 112. |
Using s, p, d, f notations, describe the orbital with the following quantum numbers: (a) `n = 2, l = 1` (b) `n =4, l =0` (c) n = 5, l = 3 (d) n = 3, l = 2 |
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Answer» (a) n = 2, l = 1 means 2p orbital (b) n = 4, l =0 means 4s orbital (c) n = 5, l = 3 means 5f orbitals (d) n = 3, l = 2 means 3d orbital |
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| 113. |
Using s, p, d notations, describe the orbital with the following quantum numbers : (a) `n=1, l=0` (b) `n=3, l=1` (c) `n=4, l=2` (d) `n=4, l=3`. |
| Answer» (a) 1s (b) 3p (c) 4d (d) 4f. | |
| 114. |
Usings,p,d,f notations, describe the orbital with the following quantum numbers. (i) `n=2, l=1` (ii) `n=4, l=0` 9iii) `n= 5 , l=3` (iv)` n=3 , l=2` |
| Answer» (a) 2p, (b) 4s, (c ) 5f, (d) 3d | |
| 115. |
In how many elements the last electron will have the following set of quantum numbers, n = 3 and l = 1?A. 2B. 8C. 6D. 10 |
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Answer» Correct Answer - C n = 3, l = 1 represents 3p orbital. Since p has three orientations `p_(x), p_(y) and p_(z)`, 6 electrons will show same quantum number values of n and l. |
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| 116. |
An orbital is described with the help of a wave function. Since many wave functions are possible for an electron, there are many atomic orbitals. When atom is placed in a magnetic field the possible number of orientations for an orbital of azimuthal quantum number 3 isA. threeB. twoC. fiveD. seven |
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Answer» Correct Answer - D When l = 3, as `m_(l)=(2l + 1) `= 7 |
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| 117. |
Principal, azimuthal and magnetic quantum numbers are respectively related toA. Size, orientation and shapeB. Size, shape and orientationC. Shape, size and orientationD. None of these |
| Answer» Correct Answer - B | |
| 118. |
When the azimuthal quantum number has the value `2`, the number of orbitals possible isA. `5`B. `7`C. `0`D. `3` |
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Answer» Correct Answer - A Each subshell of quantum number `l` contains `2l +1` orbitals. Thus, if `l = 2`, then there are `(2xx2) +1=5` orbitals. |
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| 119. |
Azimuthal quantum number defines.A. e/m ratio of electronB. spin of electronC. angular momentum of electronD. magnetic momentum of electron . |
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Answer» Correct Answer - C Conceptual question. |
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| 120. |
The correct orders of increasing energy of atomic orbitals isA. `5p lt 4f lt 6s lt 5d`B. `5p lt 6s lt 4f lt 5d`C. `4f lt 5p lt 5d lt 6s`D. `5p lt 5d lt 4f lt 6s`. |
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Answer» Correct Answer - B Sequence is correct according to Aufbau Principle. |
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| 121. |
Which of the following sets of quantum numbers represents an impossible arrangement ?A. `{:("n",,"l",,"m",,"s",),(3,,2,,-2,,1//2,):}`B. `{:("n",,"l",,"m",,"s",),(4,,0,,0,,1//2,):}`C. `{:("n",,"l",,"m",,"s",),(3,,2,,-3,,1//2,):}`D. `{:("n",,"l",,"m",,"s",),(5,,3,,0,,-1//2,):}` |
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Answer» Correct Answer - C For l=2 (d-subshell), the value of m cannot be - 3. |
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| 122. |
Identify the incorrect statement among the following.A. 4f and 5f orbitals are equally shieldedB. d-Block elements show irregular and erratic chemical properties among them-selves.C. La and Lu have partially filled d-orbitals and no other partically filled orbitalsD. The chemistry of various lanthanoids is similar . |
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Answer» Correct Answer - A 4f orbitals are more shielded than 5f orbitals since these are chooser to the nucleus. |
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| 123. |
The orientation of an atomic orbital is governed by :A. Principal quantum numberB. Azimuthal quantum numberC. Spin quantum numberD. Magnetic quantum number. |
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Answer» Correct Answer - D is the correct answer. |
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| 124. |
When cathode rays strike the surface of hard metals like tungsten, molybdenum etc. .....are produced. |
| Answer» Correct Answer - X-rays | |
| 125. |
Which orbital in each of the following pairs is lower in energy in a many-electron atom? (a) 2s, 2p (b) 3p, 3d (c) 3s, 4s (d) 4d, 5f |
| Answer» (a) `2s lt 2p` (b) `p lt 3d` (c) `3s lt 4s` (d) `4d lt 5f` | |
| 126. |
The principle quantum number, `n`, describes the_____an electron occupies.A. subenergy levelB. main enegry levelC. secondary enegry levelD. tertiary enegry level |
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Answer» Correct Answer - B It is also referred to as shell. |
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| 127. |
Which subatomic particle is not present in an ordinary hydrogen atom ?A. protonB. neutronC. nucleusD. electron |
| Answer» Correct Answer - B | |
| 128. |
Which one of the following is a correct electronic configuration of sodium?A. 2, 8, 1B. 8, 2, 1C. 2, 1, 8D. 2, 8, 2 |
| Answer» Correct Answer - A | |
| 129. |
Calculate the frequency of paritcle wave, when the kinetic energy of a sub-atomic particle is`3.79 xx 10^(-27) J`. Hint `: KE = (1)/(2) mv^(2)` `lambda = (h)/(mv) , lambda = ( v)/( v) ` `:. (v)/(v) = (h)/( mv), v= (mv^(2))/( h) = ( 2 xx ( 1)/(2) mv^(2))/( h) ` |
| Answer» ` v= 1.143 xx 10^(7) s^(-1) ( Hz)` | |
| 130. |
A Photon of red light having wavelength 600 nm has energy equal of `(h=6.6xx10^(-34)Js)` :A. `3.3xx10^(-19)J`B. `1.0xx10^(-19)J`C. `3.0xx10^(-18)J`D. `1xx10^(19)J` |
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Answer» Correct Answer - A `E=hv=h(c)/(lambda)` `=((6.610^(-34)Js)xx(3xx10^(8)ms^(-1)))/((600xx10^(-9)m))` `=3.3xx10^(-19)J`. |
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| 131. |
A neutral atom of element has 2K, 8K and 5M electrons. Find out the following : (a) Atomic No. of the element (b) Total No. of s electrons (c) Total No. of p- electrons (d) No. of protons in the nucleus and (e) Valency of the element. |
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Answer» The electronic configuration of the element with with 2K, 8L and 5M electrons will be `1s^(2)2s^(2) 2p_(x)^(2) 2p_(y)^(2) 2p_(z)^(2) 3s^(2) 3p_(x)^(1) 3p_(y)^(1) 3p_(z)^(1)` (a) Total no. of electrons `=2+8+5=15 :.` Atomic No. of the element `=15` (b) Total no. of s-electrons `=2+2+2=6` (c) Total no. of p-electron `=6+3=9` (d) Since the atom is neutral, `:.` No. of protons = No. of electrons = Atomic No. `=15` (e) Since the element has only three half-filled atomic orbitals, therefore, valency of the element `=3`. |
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| 132. |
Assertion (A): It is impossible to determine the exact position and exact momentum of an electron simutaneously Reason (R) : The path of an electron in an atom is clearly definedA. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not the correct explanation of AC. A is true but R is falseD. Both A and R are false |
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Answer» Correct Answer - C Correct R. The path of an electron in an atom is not clearly defined because its position cannot be measured with absolute accuracy. |
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| 133. |
Which of the following orbitals will have zero probability of finding the electron in the yz plane ?A. `p_(x)`B. `p_(y)`C. `p_(z)`D. `d_(yz)` |
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Answer» Correct Answer - A `p_(x)` orbital lies along x-axis, Hence, the probability of finding electron is zero in the yz plane |
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| 134. |
Orbitals are the pictorial representationof`Psi` or `Psi ^(2)` `Psi^(2)` tell about the probability of finding electron. Which of the following orbital is non directional ?A. Same number of protonsB. `2p_(x)`C. `4d_(x^(2)-y^(2))`D. `d_(z^(2))` |
| Answer» Correct Answer - 1 | |
| 135. |
Assertion : The energy of an electron is mainly determined by principal quantum number. Reason : The principal quantum number is the measure of the most probable distance of finding the electron around the nucleus.A. if both assertion and reason are correct and reason is correct explanation for assertion .B. If both assertion and reason are correct but reason in not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect. |
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Answer» Correct Answer - A Reason is the correct explanation for assertion. |
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| 136. |
Assertion : The number of electrons ejected from a metal surface depend upon the frequency of light. Reason : There is a time lag between the striking of light beam and the ejection of electrons from the metal surface.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. |
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Answer» Correct Answer - D The number of electrons ejected is proportional to the intensity or brightness of light and the electrons are ejected from the metal surface as soon as the beam of light strikes i.e., there is no time lag between the striking of light beam and the ejection of electrons from the metal surface. |
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| 137. |
In photoelectric effect, the energy of the photon striking a metallic surface is `5.6 xx 10^(-19)J`. The kinetic energy of the ejected electrons is `12.0 xx 10^(-20)J`. The work function is:A. `6.4xx10^(-19)J`B. `6.8xx10^(-19)J`C. `4.4xx10^(-19)J`D. `6.4xx10^(-20)J` |
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Answer» Correct Answer - C Work function `(E_(0))` KE-Energy of ejected electrons `(5.6xx10^(-19)-12.0xx10^(-20))J` `=4.4xx10^(-19)J`. |
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| 138. |
If photon of the wavelength 150 pm strikes an atom and one of its inner bound electron is ejected out with a velocity of `1.5 xx 10^(7) ms^(-1)`, calculate the energy with which it is bound to the nucleus. |
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Answer» Energy of the incident photon `= (hc)/(lamda) = ((6.626 xx 10^(-34) Js) (3.0 xx 10^(8) ms^(-1)))/((150 xx 10^(-12) m)) = 13.25 xx 10^(-16) J` Energy of the electron ejected `= (1)/(2) mv^(2) = (1)/(2) (9.11 xx 10^(-31) kg) (1.5 xx 10^(7) ms^(-1))^(2) = 1.025 xx 10^(-16)J` Energy with which the electron was bound to the nucleus `= 13.25 xx 10^(-16)J - 1.025 xx 10^(-16)J` `= 12.225 xx 10^(-16)J = (12.225 xx 10^(-16))/(1.602 xx 10^(-19)) eV = 7.63 xx 10^(3) eV` |
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| 139. |
A photon of wavelength `4xx 10^(-7)` m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV) (ii) the kinetic energy of the emission and (iii) the velocity of the photoelectron `(1 eV = 1.602 xx 10^(-19) J)` |
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Answer» (i) Energy of the photon `(E) = hv = (hc)/(lamda) = ((6.626 xx 10^(-34) Js) xx(3 xx 10^(8) ms^(-1)))/(4 xx 10^(-7) m) = 4.97 xx 10^(-19) J` `= (4.97 xx 10^(-19))/(1.602 xx 10^(-19)) eV = 3.10 eV` (ii) Kinetic energy of emission `((1)/(2) mv^(2)) = hv - hv_(0) = 3.10 - 2.13 = 0.97 eV` (iii) `(1)/(2) mv^(2) = 0.97 eV = 0.97 xx 1.602 xx 10^(-19) J` i.e., `(1)/(2) xx (9.11 xx 10^(-31) kg) xx v^(2) = 0.97 xx 1.602 xx 10^(-19) J` or `v^(2) = 0.341 xx 10^(12) = 34.1 xx 10^(10) or v = 5.84 xx 10^(5) ms^(-1)` |
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| 140. |
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom . Calculate the ionisation energy of sodium in kJ `mol^(-1)`.A. `494.5xx10^(-6)` J/atomB. `8169.5xx10^(-10)` J/atomC. `5.85xx10^(-15)` J/atomD. `8.214xx10^(-19)` J/atom |
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Answer» Correct Answer - D `lambda=242nm=242xx10^(-9)m` Energy required to ionise one atom of Na, `E=(hc)/lambda = ((6.626xx10^(-34)Js)xx(3xx10^(8)ms^(-1)))/(242xx10^(-9)m) =8.214xx10^(-19)J//atom` |
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| 141. |
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition, n = 4 to n = 2 of `He^(+)` spectrum ? |
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Answer» For H - like particles in general `bar(v) = (2pi^92) m Z^(2) e^(4))/(ch^(3)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = RZ^(2) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` `:.` For `He^(+)` spectrum, for Balmer transition, n = 4 to n = 2 `bar(v) = (1)/(lamda) = RZ^(2) ((1)/(2^(2)) - (1)/(4^(2))) = R xx 4 xx (3)/(16) = (3R)/(4)` For hydrogen spectrum `bar(v) = (1)/(lamda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = (3)/(4) R or (1)/(n_(1)^(2)) - (1)/(n_(2)^(2)) = (3)/(4)` which can be so for `n_(1) = 1 and n_(2) = 2`, i.e., the transition is form n = 2 to n = 1 |
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| 142. |
The first emission line of Balmer series in H spectrum has the wave number equal to :A. `(9R)/(400)cm^(-1)`B. `(7R)/(144)cm^(-1)`C. `(3R)/(4)cm^(-1)`D. `(5R)/(36)cm^(-1)` |
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Answer» Correct Answer - D For Balmer series, `n_(1)=2` and `n_(2)=3, 4, 5 .....` `bar(v)`(for first line)`=R[(1)/(2^(2))-(1)/(3^(2))]=R((1)/(4)-(1)/(9))=(5R)/(36) cm^(-1)`. |
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| 143. |
Chlorine exists in two isotopic forms, Cl-37 and Cl-35 but its atomic mass is `35.5`. This indicates the ratio of `Cl-37 and Cl-35` is approximatelyA. `1 : 2`B. `1 : 1`C. `1 : 3`D. `3 : 1` |
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Answer» Correct Answer - C If `Cl-37 and Cl-35` exist in the ratio of `1 : 3`, the average value comes out to be 35.5 |
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| 144. |
The number of radial nodes for 3p orbital is .....A. 3B. 4C. 2D. 1 |
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Answer» Correct Answer - D No. of radial nodes in 3p orbital `= n -l-1 =3 -1-1 = 1` |
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| 145. |
The number of radial nodes for 3p orbital is......A. 3B. 4C. 2D. sum of the number of protons and neutrons is same but the number of protons is different. |
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Answer» Number of angular nodes=t For `4^(th )` orbtial (n=4) and l=2 for d-orbital `therefore `Number of angular nodes=2 |
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| 146. |
The wavelength of the various lines in the hydrogen atomic emission spectrum can be related by a mathematical equation: `(1)/(lambda) = R ((1)/(n^(2))-(1)/(m^(2)))` What is the value of `n` for the Pfund series of lines?A. `6`B. `5`C. `4`D. `3` |
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Answer» Correct Answer - B Most of the lines of the Pfund series appear in the far infrared regions. |
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| 147. |
The arrangement of orbitals on the basis of energy is based upon their (n+l) value. Lower the value of (n+l), lower is the energy . For orbitals having same values of (n+l). The orbital with lower value of n will have lower energy. I. Based upon the baove information arrange the following orbitals in the increasing order of energy. (a) 1s,2s,3s,2p (b) 4s,3s,3p,4d (c) 25p,4d,5d,4f,6s (d) 5f,6d,7s,7p II. Based upon the above information Solve the question. give below. (a) hich of the following orbitals has the lowest energy `4d,4f,5s,5p` (b) which of the following orbitals has the higher energy? `5p,5d,5f,6s,6p` |
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Answer» (a) (n+l) values of 1s=1+0=1,2s=2+0=2,3s=3+0-3.2p=2xx1=2 Hence, increasing order of their energy is `1slt2slt2p lt3s` (b) `4s=4+0=4,3s=3+0=3,3p=3+1=4 4d=4+2= 6` Hence `3slt3plt4slt4d` (c) ` 5p=5+1=6,4d=4+2= 6,5d=5+2 =7 ,4f=4+3=7, 6s= 6+0= 6` Hence, `4dlt5dlt6slt4flt5d` (d) ` 5f=5+3= 8,7s=7+0=7,7p=7+1=8`. Hence, `7slt5flt6dlt7p` II. (a) (n+l) valeus of 4d `=4+2=6,4f=4+3=7,5s=5+0=5,7p= 7+1=8` Hence, 5s has the lowest energy. (b) `5p=5+1 = 6,5d=5+2=7,5f=5+3=8,6s=6+0=6 6p=6+1=7` Hence, 5f has highest energy. |
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| 148. |
If `R_(H)` represents Rydberg constant, then the energy of the electron in the ground state of hydrogen atom isA. `- (hc)/(R_(H))`B. `- (1)/(R_(H) ch)`C. `-R_(H) ch`D. `- (R_(H)c)/(h)` |
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Answer» Correct Answer - C According to Rydberg formula, `bar(v) = R_(H) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`....(i) According to Bohr model, for an electronic transition, `Delta E = - (2pi^(2) me^(4))/(h^(2)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` But `Delta E = hv = (hc)/(lamda) = hc bar(v)` `:. hc bar(v) = - (2pi^(2) me^(4))/(h^(2)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` or `bar(v) = - (2pi^(2) me^(4))/(ch^(3)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`...(ii) Comparing eqns. (i) and (ii) `R_(H) = - (2pi^(2) me^(4))/(ch^(3)) :. E_(1) = - (2pi^(2) me^(4))/(h^(2)) = - R_(H) ch` |
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| 149. |
Calculate the total number of angular nodes and radical nodes present in 3p orbital. |
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Answer» For 3p-orbital , principal quantum number, n=3 and azimuthal quantum number l=1 Number of anugalr nodes =n=1 Number of radial nodes `=n-l-1=3-1-1=1` |
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| 150. |
Calculate the wavelength of the spectral line in Lyman series corresponding to `n_(2) = 3` |
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Answer» Correct Answer - 102.6 nm For Lyman series, `n_(1) = 1` `:. bar(v) = R ((1)/(1^(2)) - (1)/(3^(2))) = 109677 xx (8)/(9) = 97490.7 cm^(-1)` `lamda = (1)/(bar(v)) = (1)/(97490.7 cm^(-1)) = 102.6 xx 10^(7) cmm = 102.6 nm` |
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