Answer:
a) 23.2 e V
b) energy of the original photon is 36.8 eV
Explanation:
given,
energy at ground level = -13.6 e V
energy at first exited state = - 3.4 e V
A photon of energy ionized from ground state and electron of energy K is released.
h ν₁ - 13.6 = K
K combine with photon in first exited state giving out photon of energy
= 26.6 e V
h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°
K + ( 3.4 ) = 26.6 e V
a) energy of free electron
K = 26.6 - 3.4 = 23.2 e V
b) energy of the original photon
h ν₁ - 13.6 = K
h ν₁ = 23.2 + 13.6
= 36.8 e V
energy of the original photon is 36.8 eV
Answer:
1.34 x 10^3 Pa
Explanation:
density of oil = 0.85 x 10^3 kg/m^3
g = 9.81 m/s^2
height of oil column = 16.1 cm = 0.161 m
Pressure on the surface of water = height of oil column x density of oil x g
= 0.161 x 0.85 x 10^3 x 9.81 = 1.34 x 10^3 Pa
Thus, the pressure on the surface of water is 1.34 x 10^3 Pa.
Answer:
x = 727.5 km
Explanation:
With the conditions given using trigonometry, we can find the tangent
tan θ = CO / CA
With CO the opposite leg and CE is the adjacent leg which is the distance from the Tierral to Sun
D =150 10⁶ km (1000m / 1 km)
D = 150 10⁹ m.
We must take the given angle to radians.
1º = 3600 arc s
π rad = 180º
θ = 1 arc s (1º / 3600 s arc) (pi rad / 180º) =
θ = 4.85 10⁻⁶ rad
That angle is extremely small, so we can approximate the tangent to the angle
θ = x / D
x = θ D
x = 4.85 10-6 150 109
x = 727.5 103 m
x = 727.5 km
Answer:
Im not really sure but Id say weather .
Explanation:
