52671 + Interview Questions in Chemistry in Class 11

1.	An element A forms a chloride which contains 29.34% by weight of chloride and is isomorphous with KCl. The molecular weight of A is
Answer» `85.49` 40 23 `137.5` Solution :WTOF chlorine = 29.34 g wt of element = `100-29.34=70.66g` `(W_(1))/(W_(2))=(E_(1))/(E_(2))` `(29.34)/(70.66)=(35.5)/("EQ wt of element")` Ecq wt of element = 85.49 A is isomorphous with KCl so valency of element `A=1` `therefore` at wt of `A=1xx85.49=85.49`

Discussion

2.	An element A crystallises in fcc structure. 200 g of this element has 4.12 xx 10^24 atoms The density of A is 72 cm^(-3). Calculate the edge length of the unit cell.
Answer» SOLUTION :`RHO=(Zxxm)/(a^3xx10^(-30))"or" a=((Zxxm)/(rhoxx10^(-30)))^(1//3)=(4xx200//(4.12xx10^24))/(7.2xx10^(-30))"PM"`=299.9 pm

Discussion

3.	An element A crystallises in fcc structure. 200 g of this element has4.12 xx 10^(24) atoms. The density of A is7.2 " g cm" ^(-3) . Calculate the edge length of the unit cell.
Answer» <P> Solution :`p = ( Z xxm)/(a^(3)xx 10^(-30)) or a = (( Z xxm)/(p xx 10^(-30)))^(1//3) = ( 4xx200 //(4.12 xx10^(24)))/(7.2 xx 10^(-30))"pm" = 299.9 "pm"`

Discussion

4.	An electronic transition from 1s orbital of an atom causes
Answer» ABSORPTION of ENERGY RELEASE of energy Both release or absorption of energy UNPREDICTABLE ANSWER :A

Discussion

5.	An electron releasing group will not stabilize which of the following groups ?
Answer» Carbocation Carbanion Free radical None of these Answer :B

Discussion

6.	An electron , practically at rest , is initially accelerated through a potential difference of 100 volts . It then has a de Broglie wavelength = lambda_(1) "Å" . It then gets retarded through 19 volts and then has wavelength lambda_(2) "Å" . A further retardation through 32 volts changes the wavelength to lambda_(3) . What is (lambda_(3) - lambda_(2))/(lambda_(1)) ?
Answer» `(20)/(41)` `(10)/(63)` `(20)/(63)` `(10)/(41)` ANSWER :C

Discussion

7.	An electron of velocity ‘x’ is found to associate with a wave. The velocity to be possessed by the neutron to have half the de-Broglie wavelength possessed by electron is
Answer» `X/(1840)` `(x)/(1480)` `(x)/(920)` `1840x` SOLUTION :`(lambdae)/(lambdan) = (m_n VN)/(m_e Ve)`

Discussion

8.	An electron or a proton which one will have a higher velocity to produce matter waves of the same wavelength? Explain it.
Answer» Solution :From de-Broglie equation, wavelength `LAMBDA = H/(mv)` for same wavelength with two different particles, (ie) electron and proton `m_1v_1 = m_2v_2` (h is constant) LESSER the MASS of the PARTICLE, greater will be the velocity. Hence electron will have higher velocity.

Discussion

9.	An electron is revolving in the 2^(nd) orbit of He^+ ion. To this if 12.1 eV of energy supplied. Then to which orbit it will be excited.
Answer» 6 8 4 2 Solution :`E=- 13.6xx(Z^2)/(n^2) EV//"ATOM"` for `He^+ (Z=2)n=2 thererore E = 0 13.6 eV` SUPPLIED ENERGY = 12.1 eV `=-1.5 eV` (i.e. energy of `6^(th)` orbit)

Discussion

10.	An electron is moving in Bohr's fourth orbit. Its de Broglie wavelength is lambda. What is the circumference of the fourth orbit ?
Answer» `(2)/(LAMBDA)` `2lambda` `4 lambda` `(4)/(lambda)` ANSWER :C

Discussion

11.	An electron is moving in Bohr's fourth orbit. Its de Broglie wavelength is lambda. What is the circumference of the fourth orbit?
Answer» `2// LAMBDA ` `2 lambda ` `3 lambda ` `3//lambda ` ANSWER :C

Discussion

12.	An electrons is moving in Bohr's fourth orbits De-Broglie wavelength s lambda.What is the circumference of the fourth orbit?
Answer» `LAMBDA` `2lambda` `3lambda` `4LAMBDA` ANSWER :C

Discussion

13.	An electron is in a 4f orbital. What possible vales for the quantum, n, l, m and s can it have?
Answer» Solution :Since the electron is in a 4F ORBITAL, the value of the principle quantum number, n = 4 For the f-orbital, the SECONDARY quantum number, l = 3 The values of the MAGNETIC quantum (m) are `-l " to " +l` including zero. Therefore, when l = 3, m has seven values, i.e., `-3, -2, -1, 0, +1, +2, and 3` For each value of m, the spin quantum number, s has two values, i.e., `s = +1//2 and s = -1//2`

Discussion

14.	An electron is in one of the 3d orbitals. Give the possible values of n, l and m_(l) for this electron.
Answer» SOLUTION :For 3D ORBITAL, `n=3, L=2`. For `l=2, m_(l)=-2-1, 0+1+2`.

Discussion

15.	An electron is allowed to move freely in a closed cubic box of length of side 10 cm. The uncertainty in its velocity will be :
Answer» a. `3.35xx10^(-4) m "sec"^(-1)` B. `5.8xx10^(-4) m "sec"^(-1)` C. `4xx10^(-5) m "sec"^(-1)` d. `4 xx10^(-6) m "sec"^(-1)` Solution :`Delta x . ""m Delta v GE h/(4PI)`

Discussion

16.	An electron in the hydrogen atom absorbs energy and jumps to the 4th orbit. Then it jumps back to the ground state in steps, i.e., from 4th to 3rd orbit, then from 3rd to 2nd orbit and finally to the ground state. If a_(0) represents the Bohr radius, the de Broglie wavelength of the electron when it moves in the 3rd orbit after absorbing some definite amount of energy will be
Answer» `a_(0)//3` `9a_(0)` `2PI a_(0)` `6 PI a_(0)` Solution :`lamda = (h)/(mv) , mv R = n (h)/(2pi) or (h)/(mv) = (2pi r)/(n)` `:. lamda = (2pir)/(n)`. But `r_(3) = a_(0) xx 3^(2) = 9 a_(0)` `:. lamda = (2pi)/(3) xx 9a_(0) = 6pi a_(0)`

Discussion

17.	An electron in the hydrogen atom absorbs energy and jumps to the 4th orbit. Then it jumps back to the ground state in steps, i.e., from 4th to 3rd orbit, then from 3rd to 2nd orbit and finally to the ground state. If the ionization potential of the hydrogen atom in the ground state is 13.6 eV, the longest wavelength of the radiation required to remove the electron from Bohr's first orbit will be approximately
Answer» `612 Å` `712 Å` `812 Å` `912 Å` Solution :Ionization energy `= 13.6 XX 1.602 xx 10^(-19) J` `=21.78 xx 10^(-19) J` `E = hv = (hc)/(lamda)` `:. lamda = (hc)/(E) = ((6.626 xx 10^(-34) Js) (3 xx 10^(8) ms^(-1)))/(21.78 xx 10^(-19) J)` `= 9.12 xx 10^(-8) m = 912 Å`

Discussion

18.	An electron in the hydrogen atom absorbs energy and jumps to the 4th orbit. Then it jumps back to the ground state in steps, i.e., from 4th to 3rd orbit, then from 3rd to 2nd orbit and finally to the ground state. If lamda_(1) is the wavelength of the line for jump of the electron from 4th to 3rd orbit and lamda_(2) that of the line for jump from 3rd to 2nd orbit, then wavelength of the line for the jump from 4th orbit to 2nd orbit will be
Answer» `lamda_(1) + lamda_(2)` `(lamda_(1) lamda_(2))/(lamda_(1) + lamda_(2))` `(lamda_(1) lamda_(2))/(lamda_(1) - lamda_(2))` `(lamda_(1) + lamda_(2))/(lamda_(1) lamda_(2))` SOLUTION :`bat(V)_(1) = (1)/(lamda_(1)) = R ((1)/(3^(2)) - (1)/(4^(2)))` `bar(v)_(2) = (1)/(lamda_(2)) = R ((1)/(2^(2)) - (1)/(3^(2)))` `bar(v) = (1)/(lamda) = R ((1)/(2^(2)) - (1)/(4^(2)))` Hence, `(1)/(lamda) = (1)/(lamda_(1)) + (1)/(lamda_(2)) = (lamda_(1) + lamda_(2))/(lamda_(1) lamda_(2)) or lamda = (lamda_(1) lamda_(2))/(lamda_(1) + lamda_(2))`

Discussion

19.	An electron in the hydrogen atom absorbs energy and jumps to the 4th orbit. Then it jumps back to the ground state in steps, i.e., from 4th to 3rd orbit, then from 3rd to 2nd orbit and finally to the ground state. The total number of lines produced in the spectrum will be
Answer» 3 6 9 12 Solution :Total no. of lines PRODUCED when the electron RETURNS to the GROUND state `SIGMA (n_(2) - n_(1)) = Sigma (4 -1) = Sigma 3 = 3 + 2 + 1 = 6`

Discussion

20.	An electron in Li^(2+) ion is in excited state (n_(2)). The wavelength corresponding to a transition to second orbit is 48.24 nm. From the same orbit, wavelength corresponding to a transition to third orbit is 142.46 nm. The value of n_(2) is:
Answer» ANSWER :5

Discussion

21.	An electron in the first shell will have ________ stability and _______ energy than an electron in the third shell.
Answer» SOLUTION :LARGER , LOWER

Discussion

22.	An electron is in one of the 3dorbitals.Givethe possiblevalues of n , l andm_(l)for thiselectron.
Answer» SOLUTION :for 3DORBITAL n= 3 .l =2 So `m_(i) = 2-1 ,-0,+ 1,+2`

Discussion

23.	An electron in an excited state of hydrogen atom absorbs 2.5 times the energy needed to unboundthe electron and moves with a wavelength of 5.416 Å (h = 6.6xx10^(-34) Js) The electron was originally in
Answer» N = 3 a=4 n=5 n=2 Solution :So `e^-` is `2^(ND)` orbit

Discussion

24.	An electron in an excited state of hydrogen atom absorbs 2.5 times the energy needed to unboundthe electron and moves with a wavelength of 5.416 Å (h = 6.6xx10^(-34) Js)The energy absorbed by the electron is
Answer» 8.5 eV 3.4 eV 68 eV 3.78 eV Solution :TOTAL ENERGY ABSORBED by ELECTRON is 3.4+ 5.1 = 8.5 eV

Discussion

25.	An electron in an excited state of hydrogen atom absorbs 2.5 times the energy needed to unboundthe electron and moves with a wavelength of 5.416 Å (h = 6.6xx10^(-34) Js) The kinetic energy of the electron is
Answer» 3.4 eV 5.1 eV 13.6 eV 10.2 eV SOLUTION :`E= hupsilon + 1/2 mv^2 , X XX 2.5 = x + (hc)/(LAMBDA)`

Discussion

26.	An electron in an atom has m = -2 value. Then (I) its 'n' value should be greater thn 2 (II) its 's' value should +1/2 (III) its 'I' value should be 2 (IV) its 'l' value shoul be greater than 1
Answer» I and II are correct II and III are correct III and IV are correct I and IV are correct Solution :`m_l = -2 , l le L N ge 3`

Discussion

27.	An electron in a hydrogen atom in the ground state absorbs energy equal to 1.5 time the minimum energy required to remove the electron from the hydrogen atom. Calculate the wavelength of the electron emitted.
Answer» Solution :Energy required to remove electron from ground state of H-atom = 13.6 eV `:.` Energy absorbed by the electron `= 1.5 xx 13.6 eV = 20.4 eV` After the REMOVAL of electron from the atom, extra energy which is converted into kinetic energy `= 20.4 - 13.6 = 6.8 eV` Thus, `(1)/(2) mv^(2) = 6.8 xx 1.602 xx 10^(-19)J` or `v = sqrt((2 xx 6.8 xx 1.602 xx 10^(-19))/(9.11 xx 10^(-31))) = sqrt(2.4 xx 10^(12)) = 1.55 xx 10^(6) MS^(-1)` `lamda = (h)/(mv) = (6.63 xx 10^(-34) KG m^(2) s^(-1))/((9.11 xx 10^(-31) kg) (1.55 xx 10^(6) ms^(-1))) = 4.70 xx 10^(-10) m`

Discussion

28.	An electron in a certain Bohr orbit has velocity 1/275 of the velocity of light. In which orbit the electron is revolving?
Answer» SOLUTION :Velocity of ELECTRON `= (1)/(275) xx 3 xx 10^(10) cm s^(-1) = 1.90 xx 10^(8) cm s^(-1)` Velocity of electron in nth orbit of H-like PARTICLES in C.G.S. UNITS is GIVEN by `u_(n) = (2pi Ze^(2))/(nh)` Putting Z = 1 for H and `e = 4.80 xx 10^(-10)" esu"` and `h = 6.626 xx 10^(-27)` erg sec, we get `u_(n) = (2 xx 3.14 xx (4.80 xx 10^(-10))^(2))/(n xx (6.626 xx 10^(-27))) = (2.18 xx 10^(8))/(n)` `:. (2.18 xx 10^(8))/(n) = 1.09 xx 10^(8)` (calculated above) `:. n = 2`

Discussion

29.	An electron having velocity 2xx10^(6)m//s has uncertainty in kineticenergy is (6.66)/pixx10^(-21) J, then calculate the uncretainty in position (in Angstrom ,Å) of the electron .[Given :h=6.60xx10^(-34) J-sec]
Answer» SOLUTION :`KE=(1)/(2) mv^(2)""V=2xx10^(6)` `d(KE)=mvdv` `dv=(d(KE))/(mv)""......(1)""But""DELTA"x"=(h)/(4pimDeltav)""....(2)` `Delta"x"=(h)/(4pim(d(KE))/(mv))"" , ""Delta"x"=(6.62xx10^(-34)xx2xx10^(6))/(4pixx6.62/(pi)xx10^(-21))=500Å`

Discussion

30.	An electron has magnetic quantum number as '-3'. Its principal quantum number is
Answer» 3 2 1 4 Solution :`m_1 = PM 3, L = 3 , N GT 3`

Discussion

31.	An electron has a speed of 500 ms^(-1) with an uncertainty of 0.02%. What is the uncertainty in locating its position ? [Give mass of electron = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) J s)
Answer» Solution :Uncertainty in SPEED `(Delta V) = (0.02)/(100) XX 500 m s^(-1) = 0.1 MS^(-1)`

Discussion

32.	An electron from one Bohr stationaryorbitcan goto nexthigherorbit
Answer» by emissionof ELECTROMAGNETICRADIATION by ABSORPTIONOF electromagneticradiation by ABSORPTION of electromagneticradiationof particularfrequency . withoutemissionor absorptionof electronmagneticradiation . ANSWER :C

Discussion

33.	An electron - donating substituent in benzene ring gives a meta product.
Answer» ANSWER :F

Discussion

34.	An electron beam on hitting a ZnS screen produces a scintillation on it. What do you conclude ?
Answer» SOLUTION :ELECTRON has PARTICLE NATURE

Discussion

35.	An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelenegth becomes equal to 1.54 Å?
Answer» Solution :The aim is to find the energy of the electron in electron volts (eV). As K.E. `= (1)/(2) MV^(2)`, therefore, first we calculate velocity of the electron to have a wavelength of `1.54 Å`. Applying de BROGLIE equation `lamda = (h)/(mv)` `v = (h)/(m XX lamda) = ((6.62 xx 10^(-34) kg m^(2) s^(-1)))/((9.1 xx 10^(-31) kg) (1.54 xx 10^(-10) m)) = 4.72 xx 10^(6) m s^(-1)` Now, K.E. `= (1)/(2) mv^(2) = (1)/(2) (9.1 xx 10^(-31)kg) (4.72 xx 10^(6) ms^(-1))^(2) = 1.01 xx 10^(-17) kg m^(2) s^(-2)` `= 1.01 xx 10^(-17) J " " ( :' 1J = 1 kg m^(2) s^(-2))` `= (1.01 xx 10^(-17))/(1.602 xx 10^(-19)) eV " " ( :' 1 eV = 1.602 xx 10^(-19) J)` `= 63.12 eV` Hence, potential required = 63.12 volts

Discussion

36.	An electron beam after hitting a nickel crystal produces a diffraction pattern ? What do you conclude ?
Answer» SOLUTION :ELECTRON has wave-nature

Discussion

37.	An electron, a proton and an alpha particle have kinetic energies of 16E, 4E and E respectively. What is the qualitative order of their de Broglie wavelengths?
Answer» `lambda_egt lambda_p = lambda_a` `lambda_p= lambda_a GT lambda_e` `lambda_pgt lambda_e gt lambda_a` `lambda_agt lambda_e gt gtlambda_p` SOLUTION :`LAMBDA = H/(SQRT(2mKE))`

Discussion

38.	An electric discharge is passed through a mixture containing 50cc of O_(2) and 50cc of H_(2). The volume of the gases formed at 110^(@)C will be
Answer» 50cc 75cc 25cc 100cc Solution :`2H_(2)+O_(2)rarr2H_(2)O` `(50)/(2)lt(50)/(1)` `implies H_(2)` is the L.R `2ml H_(2)RARR 2ml H_(2)O` `50ml H_(2)rarr50mlH_(2)O` VOLUME of `O_(2)` used = 25ml implies leftover 25ml `THEREFORE` TOTAL volume of gases = `50+25=75ml`

Discussion

39.	An electric current is passed through two voltameters connected in series and containing CuSO_4 and AgNO_3 solutions respectively. The masses of copper and silver deposited are 0.424 g and 1.44 g respectively. Find the equivalent mass of silver if that of copper is 31.75.
Answer» ANSWER :107.8

Discussion

40.	An efficiency of an engine is always ____
Answer» SOLUTION :` LT 100%`

Discussion

41.	An azeotropic solution of two liquids has a boiling point lower than either of the boiling points of the two liquids when it.
Answer» SHOWS negative DEVIATION shows positive deviation SHOW no deviation is UNSATURATED Answer :B

Discussion

42.	An average personneeds about 10,000 kJ per day. How muchcarbohydrates ( in mass) willhe haveto consume, assuming that all his energy needs are met only by carbohydrates in the form glucose ? Given that the heat of combustion of glucoseis 2900 kJ mol^(-1)
Answer» Solution :MOLAR MASS of glucose `(C_(6)H_(12)O_(6)) = 180 g mol^(-1)` . Glucose requried PER day `=( 180 // 2900 ) xx10,000 g `

Discussion

43.	An automobile antifreeze consist of 38.7% carbon, 9.7% hydrogen and remaining oxygen by weight. When 0.93g of it are vaporised at 200^(@)C and 1 atm pressure, 582 mL of vapour are formed. Find molecular formula of antifreeze.
Answer» ANSWER :`C_(2)H_(6)O_(2)`

Discussion

44.	An atomic orbital
Answer» defines exactly the POSITION of an electron relative to the nucleus has a shape dependent on the spin of the electron is always spherical in shape REPRESENTS the PROBABILITY of FINDING an electron at a certain distance from the nucleus. ANSWER :D

Discussion

45.	An atomorbitalshas n=3 .What are thepossiblevaluesof l and m_(l) ?
Answer» SOLUTION :L=0 ,1,2 so (n-1)equalvalueof l l=0than `m_(l) =0` l=1than `m_(l) = 1 ,0,+1 ` `l=2` than`m_(l)= 2 -1 0,+1+2` Thustotalvalueof `m_(i)= 1+3+5= 9`

Discussion

46.	An atom of radium combines with two atoms of chlorine to formRaCl_(2) molecules . The radioactivity of RaCl_(2) will be
Answer» as MUCH as that of same QUANTITY of Ra one THIRD of the same qunatity of Ra one HALF of the same quantity of Ra zero Answer :A

Discussion

47.	An atom of an element is 55.847 times as heavier as 1/12 mass of an atom of C^12. What is the atomic mass of the element in amu ?
Answer» Solution :Since an atom of the given element is 55.847 TIMES heavier than 1/12 MASS of an atom of `C^(12)`, the atomic mass of the given element, by definition, will be 55.847 amu.

Discussion

48.	An atom of an element contains 35 electrons and 45 neutrons.Deduce(i)the number of protons(ii)the electronic configuration for the element(iii)All the four quantum numbers for the last electron
Answer» Solution :An element X contains 35 ELECTRONS 45 electrons (i) The NUMBER of protons must be equal to the number of electrons.So the number of protons=35 (ii)Number of electrons =35 so the ELECTRONIC configuration is `1s^(2)2S^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(10)4p^(2)` (iii) The last ELECTRON i.e `5^(h)` electron in 4p orbital has the following quantum numbers. n=4,l=1,m=+1,s=-1/2

Discussion

49.	An atom of an element contains 35 electrons and 45 neutrons. Deduce. (i) the number of protons (ii) the electronic configuration for the element (iii) All the four quantum numbers for the last electron.
Answer» Solution :no. of ELECTRONS : 35 (GIVEN) no. of protons : 35 Electronic configuration `1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4p^5` last electron present in `4p_y` orbital `n = 4 l = 1 m_l = "EITHER" + 1 or -1 and s = -1//2`

Discussion

50.	An atom of an element contains 29 electrons and 35 neutrons. Deduce (i)the number of protons . (ii)the electronic configuration of the element.
Answer» Solution :(i) For a neutral atom,NUMBER of ELECTRONS = number of protons 29 electrons = 29 protons. (ii) `29Z = 1s^(2),2s^(2),2p^(6),3s^(2),3p^(6),3d^(10),4S^(1)` (The element is copper).

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

An element A forms a chloride which contains 29.34% by weight of chloride and is isomorphous with KCl. The molecular weight of A is

An element A crystallises in fcc structure. 200 g of this element has 4.12 xx 10^24 atoms The density of A is 72 cm^(-3). Calculate the edge length of the unit cell.

An element A crystallises in fcc structure. 200 g of this element has4.12 xx 10^(24) atoms. The density of A is7.2 " g cm" ^(-3) . Calculate the edge length of the unit cell.

An electronic transition from 1s orbital of an atom causes

An electron releasing group will not stabilize which of the following groups ?

An electron of velocity ‘x’ is found to associate with a wave. The velocity to be possessed by the neutron to have half the de-Broglie wavelength possessed by electron is

An electron or a proton which one will have a higher velocity to produce matter waves of the same wavelength? Explain it.

An electron is revolving in the 2^(nd) orbit of He^+ ion. To this if 12.1 eV of energy supplied. Then to which orbit it will be excited.

An electron is moving in Bohr's fourth orbit. Its de Broglie wavelength is lambda. What is the circumference of the fourth orbit ?

An electron is moving in Bohr's fourth orbit. Its de Broglie wavelength is lambda. What is the circumference of the fourth orbit?

An electrons is moving in Bohr's fourth orbits De-Broglie wavelength s lambda.What is the circumference of the fourth orbit?

An electron is in a 4f orbital. What possible vales for the quantum, n, l, m and s can it have?

An electron is in one of the 3d orbitals. Give the possible values of n, l and m_(l) for this electron.

An electron is allowed to move freely in a closed cubic box of length of side 10 cm. The uncertainty in its velocity will be :

An electron in the hydrogen atom absorbs energy and jumps to the 4th orbit. Then it jumps back to the ground state in steps, i.e., from 4th to 3rd orbit, then from 3rd to 2nd orbit and finally to the ground state. The total number of lines produced in the spectrum will be

An electron in Li^(2+) ion is in excited state (n_(2)). The wavelength corresponding to a transition to second orbit is 48.24 nm. From the same orbit, wavelength corresponding to a transition to third orbit is 142.46 nm. The value of n_(2) is:

An electron in the first shell will have ________ stability and _______ energy than an electron in the third shell.

An electron is in one of the 3dorbitals.Givethe possiblevalues of n , l andm_(l)for thiselectron.

An electron in an excited state of hydrogen atom absorbs 2.5 times the energy needed to unboundthe electron and moves with a wavelength of 5.416 Å (h = 6.6xx10^(-34) Js) The electron was originally in

An electron in an excited state of hydrogen atom absorbs 2.5 times the energy needed to unboundthe electron and moves with a wavelength of 5.416 Å (h = 6.6xx10^(-34) Js)The energy absorbed by the electron is

An electron in an excited state of hydrogen atom absorbs 2.5 times the energy needed to unboundthe electron and moves with a wavelength of 5.416 Å (h = 6.6xx10^(-34) Js) The kinetic energy of the electron is

An electron in an atom has m = -2 value. Then (I) its 'n' value should be greater thn 2 (II) its 's' value should +1/2 (III) its 'I' value should be 2 (IV) its 'l' value shoul be greater than 1

An electron in a hydrogen atom in the ground state absorbs energy equal to 1.5 time the minimum energy required to remove the electron from the hydrogen atom. Calculate the wavelength of the electron emitted.

An electron in a certain Bohr orbit has velocity 1/275 of the velocity of light. In which orbit the electron is revolving?

An electron having velocity 2xx10^(6)m//s has uncertainty in kineticenergy is (6.66)/pixx10^(-21) J, then calculate the uncretainty in position (in Angstrom ,Å) of the electron .[Given :h=6.60xx10^(-34) J-sec]

An electron has magnetic quantum number as '-3'. Its principal quantum number is

An electron has a speed of 500 ms^(-1) with an uncertainty of 0.02%. What is the uncertainty in locating its position ? [Give mass of electron = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) J s)

An electron from one Bohr stationaryorbitcan goto nexthigherorbit

An electron - donating substituent in benzene ring gives a meta product.

An electron beam on hitting a ZnS screen produces a scintillation on it. What do you conclude ?

An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelenegth becomes equal to 1.54 Å?

An electron beam after hitting a nickel crystal produces a diffraction pattern ? What do you conclude ?

An electron, a proton and an alpha particle have kinetic energies of 16E, 4E and E respectively. What is the qualitative order of their de Broglie wavelengths?

An electric discharge is passed through a mixture containing 50cc of O_(2) and 50cc of H_(2). The volume of the gases formed at 110^(@)C will be

An electric current is passed through two voltameters connected in series and containing CuSO_4 and AgNO_3 solutions respectively. The masses of copper and silver deposited are 0.424 g and 1.44 g respectively. Find the equivalent mass of silver if that of copper is 31.75.

An efficiency of an engine is always ____

An azeotropic solution of two liquids has a boiling point lower than either of the boiling points of the two liquids when it.

An average personneeds about 10,000 kJ per day. How muchcarbohydrates ( in mass) willhe haveto consume, assuming that all his energy needs are met only by carbohydrates in the form glucose ? Given that the heat of combustion of glucoseis 2900 kJ mol^(-1)

An automobile antifreeze consist of 38.7% carbon, 9.7% hydrogen and remaining oxygen by weight. When 0.93g of it are vaporised at 200^(@)C and 1 atm pressure, 582 mL of vapour are formed. Find molecular formula of antifreeze.

An atomic orbital

An atomorbitalshas n=3 .What are thepossiblevaluesof l and m_(l) ?

An atom of radium combines with two atoms of chlorine to formRaCl_(2) molecules . The radioactivity of RaCl_(2) will be

An atom of an element is 55.847 times as heavier as 1/12 mass of an atom of C^12. What is the atomic mass of the element in amu ?

An atom of an element contains 35 electrons and 45 neutrons.Deduce(i)the number of protons(ii)the electronic configuration for the element(iii)All the four quantum numbers for the last electron

An atom of an element contains 35 electrons and 45 neutrons. Deduce. (i) the number of protons (ii) the electronic configuration for the element (iii) All the four quantum numbers for the last electron.

An atom of an element contains 29 electrons and 35 neutrons. Deduce (i)the number of protons . (ii)the electronic configuration of the element.

An electron in the first shell will have __ stability and _ energy than an electron in the third shell.