TN 12TH PHYSICS - Online Test

The complex is [M(H₂O)₅ Cl] Cl₂

1000 ml of 1M solution of the complex gives 2 moles of Cl⁻ ions

1000 ml of 0.01M solution of the complex will give

[100 ml x 0.01M x 2Cl^-] / [1000 ml x 1M] = 0.002 moles of C⁻  ions

number of A ions = (N_c/8) = (8/8)=1

number of B ions = (N_f/2) = (6/2) =3

End A of the rod acts as an object for mirror and A' will be its image. By mirror formula, image distance of A is V_A.

V_A = -15cm

Also image distance of C

Vc= -20 cm

∴ The length of image = | V_A-V_C |

 = | -15-(20) | = 5cm

Resolution limit = 1.22 × λ / D

D - diameter of pupil = 3 mm

λ - wavelength = 500 nm

 Resolution limit = distance between the dots / resolving diameter

= 1mm/d

[ 1.22 × λ ] / D =  1mm / d

D = [1mm × D ] / [ 1.22 × λ ] = [ 1 x 10^-3 × 3 x 10^-3 ] / [ 1.22 × 500 x10^-9 ]

 d = 4.918 ≈ 5m

deBroglie wavelength λ ∝ 1/√V

 λ₁/λ₂ = √(V₂/V₁) = √(224/14) = √16 = 4

λ₁ = 4λ₂

λ₂ = λ₁/4

Angular momentum L = n h

For n = 4, L = 4^h = 4h /2 π =2h/ π