Answer:
d) 2Fr
Explanation:
We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².
We now evaluate the integral from r = +r to -r
W = ∫Fdr
= ∫(-e²/4πε₀r²)dr
= -∫e²dr/4πε₀r²
= -e²/4πε₀∫dr/r²
= -e²/4πε₀ × -[1/r] from r = +r to -r
W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.
Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.
So W = -2e²/4πε₀r = 2Fr.
So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr
Answer:
Worldwide Radio Communication
Explanation:
The ionosphere is important because it is through the ionosphere that world wide radio communication is possible.
Answer:
Second order line appears at 43.33° Bragg angle.
Explanation:
When there is a scattering of x- rays from the crystal lattice and interference occurs, this is known as Bragg's law.
The Bragg's diffraction equation is :
.....(1)
Here n is order of constructive interference, λ is wavelength of x-ray beam, d is the inter spacing distance of lattice and θ is the Bragg's angle or scattering angle.
Given :
Wavelength, λ = 1.4 x 10⁻¹⁰ m
Bragg's angle, θ = 20°
Order of constructive interference, n =1
Substitute these value in equation (1).
d = 2.04 x 10⁻¹⁰ m
For second order constructive interference, let the Bragg's angle be θ₁.
Substitute 2 for n, 2.04 x 10⁻¹⁰ m for d and 1.4 x 10⁻¹⁰ m for λ in equation (1).
<em>θ₁ </em>= 43.33°
As soon as you let go of it it is at its max speed because gravity is constantly pulling it down
The correct answer would be odor. Because it's sweet. Boiling shape and hardness have nothing to do with sweet and floral :)
