Suppose the horizontal velocity of the wind against a sail is 6.3 m/s parallel to its front surface and 3.8 m/s along its back surface calculate the magnitude of the force on a square meter of sail in n, given that the density of air is 1.29 kg.M^3 ?
Answers
Solution:- (C) 3.15 \times {10}^{-34} {J}/{sec}3.15×10
−34
J/sec
Energy of electron in HH-atom \left( E \right) = \cfrac{-13.6}{{n}^{2}} \; eV(E)=
n
2
−13.6
eV
\Rightarrow \; {n}^{2} = \cfrac{-13.6}{E}⇒n
2
=
E
−13.6
Given that energy of the electron in HH-atom is 1.5 \; eV1.5eV.
\therefore \; {n}^{2} = \cfrac{-13.6}{- 1.5}∴n
2
=
−1.5
−13.6
\Rightarrow \; n = \sqrt{9.067} \approx 3⇒n=
9.067
≈3
As we know that, angular momentum $$\left( p \right) is given by-
p = \cfrac{nh}{2 \pi}p=
2π
nh
whereas, h is planck's constant = 6.6 \times {10}^{-34} {J}/{sec}=6.6×10
−34
J/sec
\therefore \; p = \cfrac{3 \times 6.6 \times {10}^{-34}}{2 \times 3.14} = 3.15 \times {10}^{-34} {J}/{sec}∴p=
2×3.14
3×6.6×10
−34
=3.15×10
−34
J/sec
Hence the angular momentum of electron will be 3.15 \times {10}^{-34} \; {J}/{sec}3.15×10
−34
J/sec