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New answer posted

11 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

We know

Energy on any orbit is given by,

E n = 1 3 . 6 n 2 z 2

  E 1 = 1 3 . 6 1 2 * 9 [n1 = 1] (1)

For 3rd orbit

E 3 = 1 3 . 6 9 *   [n2 = 3]

E3 = 13.6ev

Δ E = E 3 E 1

= 13.6 – (13.6 * 9)

Δ E = 8 * 1 3 . 6 e v = p h o t o n e n e r g y

8 * 1 3 . 6 e v = h c λ

8 * 1 3 . 6 e v = 1 2 4 2 e v λ n m

λ = 1 1 . 4 1 5 * 1 0 9

= 114.15 * 1010m

New answer posted

11 months ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

3d = 0.6mm

D = 80cm

= 800mm

Path difference is given by

BP – Andhra Pradesh = Dx

d y D = ( 2 n + 1 ) λ 2   [for Dark fringe at P]

n = 0, for first dark fringe

d y D = λ 2

λ = 2 d D y

= 2 d D * d 2 [ y = d 2 , G i v e n  first dark fringe is observed on the screen directly opposite to one of the slits]

λ = 2 * 0 . 6 * 0 . 6 8 0 0 * 2

λ = 4 5 0 m m

New answer posted

11 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

In strong ligand field  C o 3 + will have t 2 g 6 e g 0 of configuration and Δ t = 4 9 Δ 0

New answer posted

11 months ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

for lamp,   P = v 2 R

R = 2 5 * 2 5 5 = 1 2 5 Ω              

l m a x = v R = 2 5 1 2 5 = 1 5 A    

I m a x = 2 2 0 1 2 5 + R = 1 5

1 1 0 0 = 1 2 5 + R        

R = 975  Ω

New answer posted

11 months ago

0 Follower 16 Views

A
alok kumar singh

Contributor-Level 10

Is bulky base, so elimination is dominating.

New answer posted

11 months ago

0 Follower 31 Views

V
Vishal Baghel

Contributor-Level 10

Truth table for Input & Output

A

B

y

1

1

0

0

0

1

0

1

1

1

0

1

1

1

0

0

0

1

0

1

1

1

0

1

Now, Truth Table for option B.

             

A

B

y

1

1

0

1

0

1

0

1

1

0

0

1

New answer posted

11 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Given absorb energy = 10.2 eV

E f = 1 3 . 6 n 2 = 3 . 4

n 2 = 4

n = 2

 

New answer posted

11 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Given I = 2r

μ 1 = 1 ( A i r )

μ 2 = 2 n

by, Snell's law

μ s i n i = μ 2 s i n r

i = 2 c o s 1 [ n 2 ]

New answer posted

11 months ago

0 Follower 16 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

11 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

V 0 = 1 5 C

V 0 = C 5 , V S = 0

f = f 0 [ C + V 0 C + V s ]

V0 observer velocity

C sound velocity

Vs source velocity

= 6 5 f 0 f 0 f 0 * 1 0 0

= 1 5 * 1 0 0

% change in frequency = 20%

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