Class 12th

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New answer posted

9 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Energy of products = 939 + 940 = 1879
ΔE + E_reactions – E_productor = 1877 – 1879 = -2MeV.
Hence it must capture a γ - ray photon of energy 2MeV.

New answer posted

9 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

X? = X? ⇒ ωL = R ⇒ 2 * 3.14 * 50 * L = 20 ⇒ L = 63.7mH

New answer posted

9 months ago

0 Follower 55 Views

A
alok kumar singh

Contributor-Level 10

  5 * 1 0 3 I 1 = 6 0

I 1 = 6 0 5 * 1 0 3 = 1 2 m A                  

Potential difference across

1 0 k Ω = 2 4 0 6 0 = 1 8 0 V            

I 2 = I 3 I 1 = 6 m A            

New answer posted

9 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Photon energy  = h c λ = 1240 e V . n m 500 n m = 2.48 e V = 2.48 * 1.6 * 10 - 19 J = 4 * 10 - 19 J

No. of photons emitted per sec =   power     photon Energy   = 100 4 * 10 - 19 s - 1 = 2.5 * 10 20 s - 1

New answer posted

9 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

n (7800) = (n + 1) 5200

New answer posted

9 months ago

0 Follower 17 Views

A
alok kumar singh

Contributor-Level 10

  I = 1 2 0 . 3 5 * 1 0 3 = 2 . 3 4 m A

V 0 = I R = ( 2 . 3 4 * 1 1 3 ) ( 5 * 1 0 3 ) = 1 1 . 7 V  

            

New answer posted

9 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

v? = 2î F? = -2?
v? = 2? F? = -2î
⇒ B? is along -k? Hence v? = 2k? ⇒ F? = 0

New answer posted

9 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Gain in K.E. =  Loss in P.E.

  K p = e V for proton

K α = 2 e V  for  α - particle

Again, Loss in K.E. =   Work against field

K p = e E S 0  for proton

e V = e E S 0

S 0 = V E

K α = 2 e E . S  for α -  particle

2 e V = 2 e E S

S = V E = S 0

New answer posted

9 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

tan 60° = (2kλ? /y? )/ (2kλ? /x? )


(√3λ? /x? ) = (λ? /y? ); √3λ? y? = λ? x?

New answer posted

9 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

  e A s u n σ T 4 4 π r 2 * A p r o j E a r t h
4 π R 2 4 π r 2 σ T 4 π r 0 2
π R 2 r 2 r 0 2 σ T 4

            

            

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