de Broglie wavelength (λ) is given by the equation
λ = h/p
where h=Planck’s constant whose value is 6.62 x 10^(−34) joule-seconds and
p = momentum of the particle(here electron)
In terms of kinetic energy(E) momentum(p) can be written as,
p=(2mE)^1/2
where m=mass of the particle.
Hence λ becomes
1 λ = h(2mE)^-1/2
Given here, E = 13.6 eV = 13.6×1.6×10^-19 joule
m(mass of electron)= 9.1×10^-31 kg
Putting these values in equation (1) we get ,
λ =0.332×10^(-9) meter
=3.32×10^(-10) meter
=3.32 Å
Potential energy at any point is (M G H). On the way down, only H changes. So halfway down, half of the potential energy remains, and the other half has turned to kinetic energy. Half of the (M G H) it had at the tpp is (0.5 x 9.8 x 10) = 49 joules.
Answer:
0.84μF
Explanation:
Charge is same through both the capacitors since they are in series. Total voltage is the sum of the voltages of the individual capacitors.. So voltage across the 2nd capacitor is 120- 90 =30 V.
Charge across first capacitor is Q = C₁V₁ = 90 x0.28 = 25.2μC
Therefore capacitance of 2nd capacitor =
C₂ = Q÷V₂ = 25.2÷30 = 0.84 μF
