Answer:
A. 2.82 eV
B. 439nm
C. 59.5 angstroms
Explanation:
A. To calculate the energy of the photon emitted you use the following formula:
(1)
n1: final state = 5
n2: initial state = 2
Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):
B. The energy of the emitted photon is given by the following formula:
(2)
h: Planck's constant = 6.62*10^{-34} kgm^2/s
c: speed of light = 3*10^8 m/s
λ: wavelength of the photon
You first convert the energy from eV to J:
Next, you use the equation (2) and solve for λ:
C. The radius of the orbit is given by:
(3)
where ao is the Bohr's radius = 2.380 Angstroms
You use the equation (3) with n=5:
hence, the radius of the atom in its 5-th state is 59.5 anstrongs
Answer:
the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
Explanation:
Given that;
weight of vehicle = 4000 lbs
we know that 1 kg = 2.20462
so
m = 4000 / 2.20462 = 1814.37 kg
Initial velocity 
= 60 mph = 26.8224 m/s
Final velocity 
= 30 mph = 13.4112 m/s
now we determine change in kinetic energy
Δk = 
m( ² - ² )
we substitute
Δk = ×1814.37( (26.8224)² - (13.4112)² )
Δk = × 1814.37 × 539.5808
Δk = 489500 Joules
we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule
so
Δk = 489500 / 3.6 × 10⁶
Δk = 0.13597 ≈ 0.136 kWh
Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
Answer:
The heat transferred into the system is 183.5 J.
Explanation:
The first law of thermodynamics relates the heat transfer into or out of a system to the change of internal and the work done on the system, through the following equations.
ΔU = Q - W
where;
ΔU is the change in internal energy
Q is the heat transfer into the system
W is the work done by the system
Given;
ΔU = 155 J
W = 28.5 J
Q = ?
155 = Q - 28.5
Q = 155 + 28.5
Q = 183.5 J
Therefore, the heat transferred into the system is 183.5 J.
Answer:
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