Answer:
1.10134 * 10⁻⁹m⁻¹
Explanation:
K = 680Nm⁻¹
μ = ?
μ = (m₁ + m₂) / m₁m₂
compound = CO
C = 12.0 g/mol = 0.012kg/mol
O = 16.0g/mol = 0.016kg/mol
μ = (m₁ + m₂) / m₁m₂
μ = (0.012 + 0.016) / (0.012*0.016) = 145.83
v = 1/2πc * √(k/μ)
ν = 1/ 2*3.142* 3.0*10⁸ * √(630/145.83)
v = 5.30*10⁻¹⁰ * 2.078
v = 1.10134*10⁻⁹m⁻¹
Explanation:
The ratio of the areas is the square of the ratio of the radii.
A/A = 3.16² = 9.99
The ratio of the volumes is the cube of the ratio of the radii.
V/V = 3.16³ = 31.6
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
the answer is a because they start on the north side
Answer:
The magnification produced by a plane mirror is +1
means then the size of the image is equal to the size of the object. If m has a magnitude greater than 1 the image is larger than the object, and an m with a magnitude less than 1 means the image is smaller than the object.
