# ► CSIR UGC NET, 2014 June 22, Question, Detailed Answer Key, Key, Answers, solutions, Explanations, Comment Here

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CSIR UGC NET, 2014 June 22, Question, Detailed Answer Key, Key, Answers, solutions, Explanations, Comment Here

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UGC CSIR NET 2014 June 22 Physics Answer Key Comment here with explanation

We can complete the detailed answer key with your help.Please comment the key only if you are sure about the answer, and also give the explanations regarding your answer.

**Download Question Paper 22 Dec 2014 Physics**

Posted by our guest writer S . If there is any content violations Report the Administrator by clicking "Report Problem".

Pls comment answers along with questions

ReplyDeletebooklet A

Delete21) 2

23) 4

25) 4

26) 3

27) 2

34) 1

35) 1

36) 3

39) 4

40) 3

50) 4

51) 4

54) 1

59) 3

61) 1

65) 4-

booklet A

DeletePART A

1) 3

4) 2

5) 2

6) 2

7) 2

9) 4

11) 4

12) 3

13) 3

14) 4

15) 3

17) 2

18) 2

19) 2

20) 2

PART B

21) 2

23) 4

24) 2

25) 4

26) 3

27) 2

34) 1

35) 1

36) 3

39) 4

40) 3

PART C

50) 4

51) 4

54) 1

59) 3

61) 1

65) 4

many answers are wrongs

Deletephysical science set c

ReplyDeletepart( a)

1-3

3-1

4-2

5-3

7-4

8-2

9-2

11-4

12-1

13-3

14-2

16-2

17-2

18-2

19-2

12 answer will be 40.5km

Deleteits 41.5 calculte it

Delete3-3 hoga 100% sure

Deleteno 12 answer is 40.5.... as in 30 minute the man covers 1.5km starting from rear end, after 500m he change his direction and moves to rear end again, after 500m he again moves towards front end of train, so after 30 min he is in front end of train, and in 30 minutes train covers a distance of 40km, so total distance between the observer and man is 40+.500=40.5km

DeleteThis comment has been removed by the author.

ReplyDeleteplease tell me cutoff marks

ReplyDelete70-75 for gen

DeleteLast tym 82 gyi thi fir iss baar itni kam kyu

Delete1 b

ReplyDelete8 d

9 b

10 b

11 c

13 d

14 b

15 b

16 b

17 d

20 b

25 c

26 c

28 c

29 b

31 b

33 b

34 d

35 c

36 b

37 c

38 d

40 c

43 a

44 d

46 c

50 d

51 d

52 d

56 a

63 d

65 b

72 d

74 d

which set no ? A B OR C

Deletewhich set no

DeleteWhat is the SET of your answer key

Deletepart-c,

ReplyDelete1-3

3-1

4-2

5-2

7-4

8-2

9-3

10-1

11-3

12-1

14-2

15-3

16-2

17-2

18-2

19-3

20-4

12-2

DeleteFourier transform and will be option 2

DeleteIn which both upper side in vertical form

You all can just go through web by just see the Fourier transform graph of cosine function

Matrices and is option four

Delete[A B]=C, [B C]=0, [C A]=B

Just do some matrix multiplication and check the result

The relation best describes the dependence of n & M

Deleten~M^4/9

X-component of velocity of electron is 4Eoa/h cut

DeleteDifferentiate given equation with respect to Kx and put the given value(Ï€/a,0,0)

And result follows

Ve= dE/h cut dKx= 4Eoa/h cut

V=-dE/h cut dkx = -4Eoa/h cut

Deletewhat will be the answer for the dependency of n(electron density) in Pressure of fermion gas?

ReplyDeleten^5/3

Deleten^5/3

Delete1/5

Delete70-75 for LS and 80-85 for Jrf in general catagory

ReplyDeleteKeshav plz write booklet code

Deletebooklet B

ReplyDeletepart a

1) 2

2) 3

3) 2

4) 1

5) 2

6) 2

7) 4

9) 2

13) 4

14) 3

16) 2

17) 4

18) 3

19) 2

20) 2

please comment answer keys of booklet code B

ReplyDeletePlease upload the answers of question booklet set "B" .

ReplyDeleteWhat is the answer of vander wall equation ' s question???

ReplyDeleteI think a/27b2

it is 3b

DeleteThis comment has been removed by the author.

Delete3b

Delete3b

DeleteAnd answer of probability within the bohr radius is

ReplyDelete0.32

Delete0.32.calculate groundstate probability with R10=2*a0^(-3/2)*exp(-r/a0).

Deleteu wil find.

the reverse of this question was asked on june 2011

DeltaC/C=?

ReplyDeleteIt is 0.14

DeleteSince DelC/C=√{(0.01)^2+(0.01)^2}

Mr. Rajesh, can you explain this equation?

DeleteDid you notice that the error of mass is in milli gram.

The answer is 0.28% given in csir official website.

SET C

ReplyDeletePart A

1) 3

3) 1

4) 2

5) 3

6) 1

7) 4

9) 3

10) 3

12) 1

13) 3

14) 2

16) 1

17) 2

18) 2

Part B

23) 1

24) 4

27) 2

29) 4

31) 1

33) 2

34) 4

42) 3

43) 3

Part C

46) 4

50) 2

59) 4

61) 4

64) 1

72) 4

CODE C:

DeletePART B:

21) 1

23) 1

26) 2

27) 2

28) 3

29) 4

30) 3

34) 4

36) 3

38) 1

42) 2 (not zero bcaz Z exist)

43) 3

PART C:

46) 4

47) 4

51) 4

55) 3

61) 4

62) 4

64) 2

65) 1

B-CODE c:

DeletePART B:

21) 1

23) 1

26) 2

27) 2

28) 3

29) 4

30) 3

34) 4

36) 3

38) 1

42) 2 (not zero bcaz Z exist)

43) 3

PART C:

46) 4

47) 4

51) 4

55) 3

61) 4

62) 4

64) 2

65) 1

What the answer of booklet- c . Question no. 61. V(x)= -§ (x) and ¥(x)=A e (a!x!). And H=b x2. Find ground state energy.

ReplyDeleteSet A

ReplyDeleteSection A

1. 3

2. 3

3. 2

4. 2

5. 2

6. 2

7. 3

9. 1

10. 1

12. 3

16. 2

17. 2

18. 3

19. 4

20. 2

Section B

25. 4

26. 3

27. 2

29. 4

31. 3

32. 4

34. 1

35. 2

36. 3

37. 1

38. 1

39. 2

40. 3

41. 1

Section C

52. 3

53. 2

59. 4

60. 1

61. 1

63. 4

65. 4

68. 2

69. 4

70. 4

71. 1

72. 2

73. 4

are u sure b<-1for stable and unstable points

Deletethink carefully qn no 10.

Deletethe answer should be 4. not 1

ans of 31-2,32-1

DeleteBooklet B

ReplyDeleteQ22 Asystem can have 3 energylevels E=0,+E,-E. The level E=0 is double degenerate.........

ANS - Option 4

n^2/3

ReplyDeleteplease apload answer key of set -c

ReplyDeleteplz give the answer keys of booklet series b for pfysical sciences june 2014

ReplyDeleteThis comment has been removed by the author.

ReplyDeletebooklet B

ReplyDelete1.2

3.2

4.3,

5.2,

6.2,

9.2,

10.1,

11.3,

13.4,

14.2,

15.3,

16.2,

17.3,

18.1,

19.3,

22.4,

24.2,

26.3,

28.1,

29.3,

31.1,

32.1,

33.2,

34.4,

35.4,

37.2,

38.2,

43.1,

44.3,

46.3,

48.3,

49.4,

51.4,

52.1,

57.4,

63.4,

64.4,

65.4,

67.2,

69.1,

74.4

i'll be happy to see your suggestions...

i feel answer for 10 is option 4

Deletebook (b)

Delete22.4

23.3

24.2

25.1

26.3

29.3

31.1

32.1

33.2

34.4

35.4

37.2

38.2

40.3

41.3

42.3

43.1

44.3

45.1

46.1

48.3

49.2

51.4

52.3

53.3

55.1

56.1

57.4

62.2

63.4

64.1

65.4

66.2

67.2

68.1

69.4

71.1

72.4

74.4

Fermi Pressure is number density times fermi energy i.e. Equivalent to n to the power five by three since fermi energy is itself n to the power two by three

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteand 17 is 2

ReplyDelete19 - 2

ReplyDeletePlz give explanation of the aswer

ReplyDeleteSeries ABCD answer is K

ReplyDeleteDerivate twice with respect to V and set it equal to zero and ans will be 3b

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteThis s nuclear physics question. Erot= (h cross)square/2I*J(J+1). Delta J =2. Next excited state s 4. Ans 310kev

DeleteYes you are right I also now referred basic nuclear physics by BN srivastava

DeleteWhere it is given the excitation energy could be predicted by E I'/E l=I'(I'+1)/l(l+1)

So that I=0;2;4;6 of even parity

Ratio E4/E2=10/3

E6/E2=7

E8/E2=12

I will have even parity only if the body has a plane of symmetry and its intrinsic angular momentum is zero

Energy of photon will be share mass of higgs minus square mass of Z boson dived by twice the mass of higgs i.e. Equal to 30.1 MeV so its ans will be 30 MeV

ReplyDeletePlease amend to read share as square

ReplyDeleteEnergy ground state will be 3 pi square h cut squareby 2 m a square i e ground level all quantum level is equal to one so share of each add will give 3

ReplyDeleteBut the given Potential is not of a paricle in 3-D box.....

Deletebooklet 2

ReplyDelete22...2

28...1

29...4

31...1

33...3

34...4

34...4

37...2

39...4

52...2

53...3

55...3

58...1

63...4

65...4

71...1

72...2

what is cut off for obc

ReplyDelete85 to 90 for JRF

Delete75 to 85 for LS

Area of triangle ABC/OST = 225

ReplyDeleteSince area of OST = 1/2(0.2)(0.2)=0.02( by similarity property check the side length ratio of triangle OPB and dotted triangle inside triangle TBR(let some point R after Pans meets T)

AREA OF ABC=9/2

Triangular series and is 80 ( multiply all and subtract it after adding all)

ReplyDeleteMinimum no of tiles and 12 because only one tile of size 5 unit can fit into it in any directing so it will be1+6+5=12

ReplyDeleteWeight of paper packet ans is 2.4 kg

ReplyDeleteLength of longer side of trapazium is 10 cm since put two vertically and one inverted between both and you will get it

ReplyDeleteHeight of tree will be (10+17=27 cm) since angle between height and horizontal distance is tan 45°

ReplyDelete10+7 = 17

DeleteSeries power of 17 and is 75 since after each interval of four you will get the sane 3 digut

ReplyDeleteTriangular series and is 80 since multiply each one and subtract after adding each

ReplyDeleteTime gap between min and hour hand is 12min since hour hand makes1/2 degree and min hand makes 6 degree I.e 5and 1/2 degree is varying every min so

ReplyDeleteNo of shirts bought is 30

ReplyDeleteRatio of weight to the square if the height is constant

ReplyDeleteProportionality can have an additive constant.therefore the ratio is constant only when the hight and weight is zero at the age zero

DeleteRounded off chicory percentage is 14

ReplyDelete197o315 ....

ReplyDeleteAns is 4

Relation between RD

ReplyDeleteRD of group > RD of group B

Prismatic crystal and 72 (4×16+8=72(

ReplyDeleteABBCCD in it AD is a perfect square

ReplyDelete1/sin^2(z) contour integration |z|=1/2

ReplyDeleteMinimum no. Of tikes is 12 because only one tile if size 5 will fit into given size and sixsize2unit and five size one unit so total if 12

ReplyDeleteOps amp when working as voltage follower then it matches high impedance source to low impedance source

ReplyDeleteThis manely working as buffer for logic circuit

It is of two type

1. Current buffer

2. Voltage buffer

Its gain is unity I.e maximum transfer of source to load.

This comment has been removed by the author.

ReplyDeleteaccording to the given potential this is 3D spherical box.

Deletethe GS energy pi^2 h(cut)^2/2ma^2 ------------griffith(2nd edition) page 153 example 4.1

according to the given potential this is 3D spherical box.

Deletethe GS energy pi^2 h(cut)^2/2ma^2 ------------griffith(2nd edition) page 153 example 4.1

Graph energy density versus wavelength at temp is and third option in which wavelength shifts towards the left side as temp increases

ReplyDeleteRefer quantum mechanics zeitilli where graph given is on energy density and frequency where it shifts towards right side on increase of temp

To get the oscillation RC network should give total of zero degree phase shifts with input I.e 360° so it requires 12 circuits as each one giving 30° shift

ReplyDeleteSir,

DeleteWe have used CE configuration here which add 180 phase shift. Do you think it reduces the requirement of RC circuits to 6 ??

Plz reply

6

Deleteya u r right

DeleteThermodynamic equilibrium then value of psi is equal to derivative of F with respect to psi and make it equal to zero so and is +/-(a/3b)^1/4

ReplyDeleteEffective capacitance is unaffected

ReplyDeleteComposite wave function for fermion is even after ignoring the spin there is at least two quantum no is required to give the sum of square is equal to 5.so option first

ReplyDelete2/a(sin(Ï€x1/a)sin(2Ï€x2/a) - sin(2Ï€x1/a)sin(Ï€x2/a))

Hamiltonian is p ^2/2(m-lambda q)

ReplyDeleteSince p=del H/del q dot

H=sum of p into q dot minus L

After calculation option 4 is matching ...is it wrong ?

DeleteHamiltonian is function of (p, q , t)

DeleteIn or option it is function of p & derivative of q so please check

The line integral is equal to surface integral of curl of given vector into ds

ReplyDeleteSince here curl is zero so and will be zero

P5(-1)= -1

ReplyDeleteFirst put the value of x=-1 and n=5 and expand both sides of power series and equate the co efficient so and will come as -1

Image problem speed if charged particle first evaluate potential then calculate velocity by using 1mv^2 formula

ReplyDeleteVelocity is proportional to square root of distance from image charge

solving we get d/5

DeleteSo and is d/4

ReplyDeletePlease do not write options give correct explanations so that hardel 75questions will be covered in few posts then in reply give your justification if you found some mistakes

ReplyDeleteEigen value of matrix is -7;0;+7

ReplyDeleteSince determinant is zero

Sum of principal diagonal element is zero

This comment has been removed by the author.

ReplyDeleteif u solve its double derivative and put value then ans is 1

Deleteare you sure the answer is 1/2 check it is 1

DeleteThis comment has been removed by the author.

DeleteIntegral of square root of X between zero and one will be 0.657

DeleteHere h=1-0/4=0.25

Yo=√Xo=0

Y1=√X1=√0.25

Y2=√X2=√0.5

Y3=√X3=√0.75

Y4=√X4=√1

Now

I=h/3[Yo+Y4+2Y2+4(Y1+Y3)]

=0.25[o+1+√2+2+2√3]/3

=0.25/3[3+1.414+3.64]

=0.25/3[0.7.878]

=2.626/4

=0.656

Charge density will be + epsilon zero phi zero by e into r square

DeleteBy using Poisson distribution for potential in spherical polar coordinate

1/r^2 del by del r( r^2 del phi by del r)= - rho by epsilon zero

Result is positive value as option one

Magnetic vector potential answer is option four

Delete-√c square t square minus a square. To. +√c square t square minus a square

T dz/(a square+ z square)^ 1/2

Kindly refer Griffith electrodynamics fourth edition page no. 447

I.e option four MU zero K/ 4Ï€ along Z

Deletecould u please tell the page number for 3ed griffith....or at least the chapter?

DeleteBKLET B KEY

ReplyDelete1B

2C

4C

5B

6B

7D

8A

9B

10D

11D

12B

13D

14B

15C

16B

17D

19A

PART B

22B

24D

25A

26B

28A

29C

30D

31A

32A

33B

34D

35D

36C

37B

38B

39D

40C

42C

43C

44B

45B

PART C

46C

50B

52C

53D

54D

55C

56A

57C

58A

63D

67D

69D

72A

74C

is the ans for q.54 is option D. please let us know the reference.

DeleteV=Vo/1-c.v/c^2 which is non relativistic formula

ReplyDeleteWhen source of light is moving radially away from the observer. V=Vo√(c-v/c+v) I.e Vo>V and non relativistic freq = Voc/c+v

If source of light is moving radially outward V=Vo√c+v/c-v I.e V>Vo and non relativistic freq=Voc/c-v

Invade light source moving transversely with respect to the observer c.v=0

I.e V=Vo√(1-v^2/c^2)

Non relativistic freq=Vo

Doppler effect does not exist in non relativistic .

In relativity V=Vo√(1-v^2/c^2)/1-c.v/c^2

So and will be f(1-v^2/c^2)^1/2

The period of oscillation of simple pendulum under gravity is given by T=2Ï€√(L/g)

ReplyDeleteHere additional acceleration is given horizontal so it will add vectorially with vertical acceleion and its resultant magnitude will be equal to 2g

So time period will be T/√2 and will make 60° with vertical

plz give all answers from Part C,

DeletePlease someone provide answers for question no. 47 and 67 of booklet A.

ReplyDeleteu can verify easily

Deletex(t)= (A+Bt) exp(-t)

ReplyDeletefirst initial condition gives A=0 ,

second initial condition gives B=1

so, x(t) = texp(-t)

now for maximum differentiate the solution

-t*expt(-t)+exp(-t)=0

exp(-t) * (1-t)=0

so t=1

A spectral line due to a

ReplyDeletetransition from an electronic

state p to an s state splits

into three zeeman lines in

presence of a strong

magnetic field. At

intermediate field strengths

the number of spertral line is

1. 10

2. 3

3. 6

4. 9

Answer with reason

J

ans will be 6

Deletemost of the answers are sure to be correct

ReplyDeleteRe(sin(X+iY))= sinXcos(iy)+cosXsin(iy)

ReplyDeleteCos(iY)= Cos h y ,

Sin(iY)= i Sin(iY)

So its real part is equal to SinXCosh Y~phi

Del phi by del X = CosXCosh Y

Del square by del X square= - Sin X Cosh Y

Del phi by del Y = SinX Sin h Y

Del square by del Y=SinX Cos h Y

Del square phi by del Z square= - Cos(z-vt)

Del square phi by del t square/v^2= - Cos(z-vt)

So and is option no. One in which

1/ v^2 del square phi by del t square is equal to ( del square by del X square plus del square by del Y square plus del z by del square) phi

The mass of meson will be 283MeV/C^2

ReplyDeleteFirst calculate the separation between quark and anti quark

Derivate the given equation with respect to r and put it equal to zero, it will give r equal to (b/a)^1/2

& corresponding potential at that separation will be equal to 2√ab

Put the values and it will come as 282.8MeV which the energy by which meson is bounded so its mass will be 283 MeV / C^2 option 2

plz explain Q.52of code A

DeleteEquate dx/dt =0 you get the value of b u can calculate x= -1 it is classical mechanics gupta kumar

Deletecentral force problem stable orbit for circle unstable hyperbola may be option (1)

Born approximation ans will be option one

ReplyDelete-4m (bits)(MU) by h cut square ( b square plus mu square) ^2

F(thita)=- 2m/ h cut square integral ( zero to infinity) (bits) V(r) Sin qr/qr . r^2 .Dr

Her V(r)= (bita) e to the power minus mu r

Use e to the power (iqr)

& integral of r to the power n into e to the power minus at is equal to factorial n by alpha to the power (n+1)

Then equate each side for imaginary part which is and as option one

Dear Rajesh may I know answer of d question to find d scattering amplitude??? Just give d right option

ReplyDeleteBooklet C q.45 Ans.1.6 since CE amp will provide phase shift 180 degrees and 6 RC network provide 180 degrees phase shift so total phase shift would be 0 or 360 degrees which is required for oscillation.

ReplyDeleteFriends please give correct explanation of question no. 25 and 56 of booklet b..

ReplyDeletePlz answer of Bcc lattice wavefunction ques 56 booklet b and ground state energy in 3D booklet b quest no 25

ReplyDeletenet-question-paper-22dec2014-physics.pdf

ReplyDeleteenergy eigenvalue of wave function Asin^3(Ï€x/a) is 9/10(Ï€^2hcut^2/ma^2)

ReplyDeleteuse sin3x=3sinx-4sin^3x

normalize the function it will give A=16/5a

integrate between zero to a for obtained function.

refer Griffith 2.36 problem

Plz what is the answer for part C first order correction to the ground state?

ReplyDeletefirst order correction to ground state us option 4

ReplyDeletescattering amplitude is

ReplyDelete-4mbu/h^2(b^2 u^2)^2

option one

random walker ans is( 1/2)^4=1/16

ReplyDeleteno it is 3/8

Deletethat is not correct answerasir

Delete6/16 or 3/8 is the correct solution in random walk problem refer solutions of rief book

I think there are only 4 cases (LLRR,LRLR,RLRL,RRLL).so fans shld b 4/16=1/4.plz explain other two cases if there r.

Deletefirst order correction is

ReplyDeleteb/2a^2

use integration of x^n.exp(-ax)= factorial n by a to the power (n 1)

on normalizing the value of A^2=2a

This comment has been removed by the author.

ReplyDelete(del P) (Del r)= h cut

ReplyDeleteas we have r = a zero so

del p= h cut / a zero

RC network n CE combination will produce oscillation with 6 RC network .thanks for ur attention .

ReplyDeletein normalized wave function is y expv[_ mW/2h( 2x^2 y^2)]

ReplyDeletedifferentiate w.r.t x and y partially twice and use two dimensional SchrÃ¶dinger equation

so option first will be the correct choice

rajeshji , what is the answer of q. 28 70 73 of booklet series C

DeletePLEASE GIVE THE ANSWER OF QN 72,69,65,62,51,45,36,30,28 OF SET A

DeleteDid anyone try the sum on fractional error in measuring concentration?

ReplyDeleteIts 39 in booklet code C.

Please ans which option is true for horses,donkeys, monkeys part A question.

ReplyDeletegenerating function

ReplyDeleteans option second q^2 P

Q= q^2

P=p/2q

time independent generating function will satisfy the condition of Q=del F /Del P

p = del F/del q

check it is satisfied by

q square P

just write p = 2qP ( which is give as P = p/2q)

#given time derivative is a function of x

ReplyDelete#put it equal to zero

# equate the values of x

#evaluate derivative of function x at those fixed points

#if you find it is greater than one then it represent the unstable condition

# if it is less than one it shows stable condition

# in given option best match is b is less than minus one I.e option third

What is the answer for transformer primary and secondary current problem, SET-B (the uploaded question) question Number 31?

ReplyDeleteAlso explain the answer.

ans. b.apply Faraday's law.current in secondary coil will b in such a direction that it opposes the change in current in primary coil.

DeletePlease give the answers of QUESTION NUMBERS 30,36,45,51,62,65,66,69,72 of Question Booklet ; A

ReplyDeletewhat is the answer key of question numbers 30,36,45,5162,65,66,69,72 of SET A

ReplyDeletedependence of viscosity on M ....Is it 4/9 or 3?? Plzz reply...

ReplyDeleteI think its 3. First have a look at this link:

Deletehttp://en.wikipedia.org/wiki/Mark%E2%80%93Houwink_equation

Now let a denote the dependence of eta on M then we can write:-

eta = k * M^a where k is a constant

Then we get

eta1/M1^a = eta2/M2^a

So, a = Log[eta1/eta2]/Log[M1/M2]

If you plugin any two pairs of data from the given list into the above eqn you always get nearly equal to 3

Hi All,

ReplyDeleteCODE B:PART C: Q NO:46. here my doubt is, we need to consider average energy(total energy) or Ground sate energy, please clarify my doubt.

Poisson distribution result is option first {C2,C3}=C1

ReplyDelete{C3,C1}=C2

use {A,B}= (del A/del x1)(del B/del p1) (del A/del x2)(del B/del p 2) (del A/del x3)(del B/ del p 3) -(del A/delp 1)(del B/del x1)-(del A/del p 2)(del B/ del x 3)-(del A / del p3)(delB/del x3)

which is satisfied by option first

Poisson distribution result is option first {C2,C3}=C1

ReplyDelete{C3,C1}=C2

use {A,B}= (del A/del x1)(del B/del p1) (del A/del x2)(del B/del p 2) (del A/del x3)(del B/ del p 3) -(del A/delp 1)(del B/del x1)-(del A/del p 2)(del B/ del x 3)-(del A / del p3)(delB/del x3)

which is satisfied by option first

The Gibbs potential answer is zero

ReplyDeleteuse 1/T= (del S / del U) at constant V

P/T = ( del S / del V) at constant U

put it in given equation of G

you will find result as

G = U- TS TS/4=U- 3TS/4

= 0(on putting the values of T and S from above)

plz comment 49 and 65 Q.no's of bcode A

DeletePlz sir answer me the question of No.-66,73 of Set-C.

DeleteHow much does the total angular momentum quantum number J change in the transition of Cr (3d6) atom as it ionizes to Cr2+ (3d4) ? [Set-C, Q-73] plz answer it someone.

ReplyDelete1. Increases by 2

2. Decreases by 2

3. Decreases by 4

4. Does not change

the total angular quantum number does not change as the value of L is same for both state and spin is also same in both state

ReplyDeleteA particle of mass m in three dimensions

ReplyDeleteis in the potential--for this Q csir ans is option 1, please share me the scenario or reference book or URL for conversion from 3d to 1d at V(r)=0 and r<a (here r is radius if i am correct ?)