A hydrogen atom in the n=7 state decays to the n=4 state. The wavelength of the photon that the hydrogen atom emits is 4592.59nm.
The Energy of photon is the energy possessed by a photon when it moves from a high energy level to a low energy level. It emits a photon of a certain wavelength. The following relation can be used to find out the relation between the energy levels and the energy possessed:
E = 13.6 × Z² (1/n₂² - 1/n₁²) eV
where, n₁ is the initial energy level i.e. n₁ =7
n₂ is the higher energy level i.e. n₂ = 4
E is the energy possessed
Z is the atomic number, Z = 1 for H-atom
Subsituting in above equation,
E = 13.6 (1/16 - 1/49) eV
E = 0.27 eV
We know that,
E = hc / λ
where, h is Planck constant
c is speed of light
λ is wavelength
On subsituting,
0.27 eV = 1240/ λ
⇒ λ = 4592.59 nm
Hence, the wavelength of photon emitted by Hydrogen atom is 4592.59nm.
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Charles Law
Explanation:
Step 1:
It is given that the original volume of the gas is 250 ml at 300 K temperature and 1 atmosphere pressure. We need to find the volume of the same gas when the temperature is 350 K and 1 atmosphere pressure.
Step 2:
We observe that the gas pressure is the same in both the cases while the temperature is different. So we need a law that explains the volume change of a gas when temperature is changed, without any change to the pressure.
Step 3:
Charles law provides the relationship between the gas volume and temperature, at a given pressure
Step 4:
Hence we conclude that Charles law can be used.
Answer:
300 cos 30 = 40 a + 40 * .2 * 10
Total force = mass * acceleration + frictional force
260 = 40 a + 80
a = 180 / 40 = 4.5 m/s^2
Check:
15 a + 15 * 10 * .2 = T acceleration of 15 kg block (assuming a = 4.5)
T = 15 (4.5) + 30 = 97.5 force required to accelerate 15 kg block
260 - 97.5 = 162.5 net force on 25 kg block
162.5 = 4.5 (25) + 25 * 10 * .2
162.5 = 112.5 + 50 = 162.5
4.5 m/s^2 checks out as correct
The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.
<h3>How to determine the friction factor</h3>
Using the formula
μ = viscosity = 0. 06 Pas
d = diameter = 120mm = 0. 12m
V = velocity = 1m/s and 3m/s
ρ = density = 0.9
a. Velocity = 1m/s
friction factor = 0. 52 × 
friction factor = 0. 52 × 
friction factor = 0. 52 × 0. 55
friction factor 
b. When V = 3mls
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 0. 185
Friction factor 
Loss When V = 1m/s
Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter
Head loss = 0. 289 × 
× 
× 
Head loss = 1. 80 × 10^8
Head loss When V = 3m/s
Head loss = 
× 
× 
×
Head loss = 5. 3× 10^8
Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.
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