The equivalent magnetic moment of an electron revolving around a nucleus

An electron in an atom revolves around nucleus in an orbit of radius 0.5 Å. Calculate the equivalent magnetic moment, if the frequency of revolution of electron is 10¹⁰MHz

CBSE – AIPMT – 2011 de Broglie wavelength associated with electron

Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100 km then the de-Broglie wavelength associated with the electrons would

(a) Increase by 4 times

(b) Increase by 2 times

(c) Decrease by 2 times

(d) Decrease by 4 times

What is the force between two small charged spheres having charges of 2 × 10–7C and 3 × 10–7C placed 30 cm apart in air?

Solution :

Given :

a)      q ₁ = 2 × 10 ^{– 7} C

b)      q ₂ = 3 × 10 ^{– 7} C

c)       r   = 30 cm = 30 X 10 ^{– 3} = 3 X 10 ^{– 2}

d)      ε ₀ = 8.854 × 10 ^{– 12} C ² N ^–1 m^{– 2}

Formula : 

The force between two small charged spheres can be given as

F = [ 1 / ( 4 π ε₀ ) ] [ ( q₁ q₂) / ( r² ) ]

Therefore,

F = [ 1 / ( 4 X 3.14 X 8.854 × 10^–12 ) ] { ( 2 × 10^–7 X 3 × 10^–7) / [ ( 3 X 10^{– 2})² ] }

F = [ 1/(1.11 × 10^–10 ) ] [ ( 6 × 10^–14 ) / ( 9 X 10 ^–4 ) ]

F = [ 9 X 10 ⁹ ) ] [ ( 6 × 10^–10 ) / ( 9 ) ]

F = 60 N

The force between two small charged spheres having charges of 2 × 10^–7C and 3 × 10^–7C placed 30 cm apart in air is 60 newton 

A planet of mass m moves in the inverse square central force field of the Sun of mass M . If the semi-major and semi-minor axes of the orbit are a and b , respectively, the total energy of the planet is:

          (a)    - ( G M m ) / ( a + b )

          (b)   - ( G M m ) [ ( 1 / a ) + ( 1 / b ) ]

          (c)    – [ ( G M m ) / a ] [ ( 1 / b ) - ( 1 / a ) ]

          (d)   - ( G M m ) / [ ( a – b ) / ( a + b )² ]