Jee 2014 Booklet1 Hwt Solutions Atomic Structure

Vidyamandir Classes

Solutions to Home Work Test/Chemistry Atomic Structure

HWT - 1

1.(C)

In lightest nuclei there is only one proton and zero neutrons and mp  (1876)me

4.(A)

Is and 25 are considered degenerate orbitals.

5.(A)

Number of photoelectrons is proportional to intensity if incident beam only if frequency of light is greater than threshold frequency.

7.(B)

K . E max  h  h0 K.E max 6.63  1019 J   1015 Hz h 6.63  1034 J H 2  1

  0  8.



For Bohr’s theory, only one proton should be resent.

Atomic Structure 1.(C)

0  (2  1)  1015 Hz  1015 Hz

HWT - 2

 1 1  E  13.6    ev 2 (n 2 )2   (n1 ) Put n 2   And n1  1

2.

Fact

3.

K.E.  eV

[V : Change in potential]

1 m2  eV 2

(e : charge]



V

2eV m

h h h   p mV 2m(K.E)

4.



5.

In n = 4 we have  16 orbitals

8.

Calcium has z = 20

10.





1s2 2s2 2p6 3s2 3p6 4s2

z = 28

for nickel



Ni  1s2 2s2 2p6 3s2 3p6 4s2 3d8



1 K.E

Ni 2  1s2 2s2 2p6 3s2 3p6 4s0 3d8 3d8  

2 unpaired e s

VMC/Atomic Structure

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HWT-Solutions/Chemistry

Vidyamandir Classes Atomic Structure 1. 2.

HWT - 3

h mV Shortest wavelength is given when largest energy change is there.  for H it is n = 1 to n   Use  

And for longest wavelength, look for smallest energy charge

He2 n = 3 to n = 2. Let wavelength by y.



 1 1 1   R  (1)2  2   R x ()2   (1) 1  R  (2)2 y

x 5  y 9

 8.

 1 1  (5)  2  2  R 4 49 (3)   (2) 

y

9x 5

Radial nodes are given by : n    1

Atomic Structure

HWT - 4

1.

Hydrogen atom has 1 proton and 1 electron. ionized hydrogen atom has only 1 proton. 

5.

r

n2 z

6.

Angular Momentum =

8.

In module

9.

In module

nh h    n   :  2  2 

Atomic Structure 1.

HWT - 5

k(z = 19) we have 1s2 2s2 2p2 3s2 3p6 4s1 Outermost e  is 4s1 1 1 4, 0, 0, + or 4, 0, 0, . 2 2 h Use   [p  mv] 5. p



2.

T

1 n

6.

Splitting of spectral lines in electric field is Stark effect. Splitting of spectral lines in magnetic field is leeman effect.

7.

Fact

9.

Isobars have equal atomic mass [protons + neutrons] Istotoes have equal atomic no. [protons] Isotones have equal neutrons.

Atomic Structure

HWT - 6 nh 2

1.

for d orbital   2

4.

Angular momentum 

5.

Fact

6.

Fact

7.

When ‘n + e’ values lashes we look for value of n and smaller the nature of n larger is the stability.

8.

Isotones have equal neutrons.

VMC/Atomic Structure

10.

 1 1   Use E    2 (n 2 )2   (n1 )

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HWT-Solutions/Chemistry

Vidyamandir Classes Atomic Structure

HWT – 7

1.

Hund’s rule is niolated in 1st case.

6.

Use K . E max  hv  w 0

7.

Spherical nodes = n    1

9.

Use Aufbav’s principle to know the configuration

Atomic Structure 1.

4.

6.

HWT - 8

These transmissions belong to lyman series Uv rays 

 2 unpaired e s

Ni 2 has 3d8 Energy of photon 

1 

hc   as E     

Atomic Structure 2.

Add all e s to get atomic number here

3.

z = 21

HWT - 9

1s2 2s2 2p6 3s2 3p6 4s2 3d1 4.

12 ‘p’ electrons

5

8.

Fact

10.

 1 ‘d’ electon. h h Use    p mv 3d1

Atomic Structure 1.

HWT - 10

Cl (z = 17) 1s2 2s2 2p6 3s2 3p5



n=3 e=1 m could be  l or.

4.

Fact.

5.

3s, 3p, 3d  1 + 3 + 5 = 9 orbitals

9.

Definition of Hund’s rule.

VMC/Atomic Structure

25

HWT-Solutions/Chemistry