The frequency of the photon emitted by a hydrogen atom is [tex]26.2*10^-^4 hz[/tex]
What are Photons?A particle representing a quantum of light or other electromagnetic radiation. A photon carries energy proportional to the radiation frequency but has zero rest mass.
In the case of light, the frequency, symbolized by the Greek letter nu (ν), of any wave equals the speed of light, c, divided by the wavelength λ:
v = c/λ
Since the wavelength λ is in the bottom of the fraction, the frequency is inversely proportional to the wavelength.
v = c/λ
v= [tex]\frac{3*10^8}{434*10^-^9} = 6.91*10^1^4 Hz[/tex]
By using Rydberg's formula,
1/λ [tex]= Rz^2(\frac{1}{n_1^2} - \frac{1}{n_2^2})[/tex]
1/λ = [tex]109677 *(\frac{1}{4^2} - \frac{1}{6^2}) = 3808.22 cm^-^1[/tex]
λ = [tex]26.2*10^-^4 hz[/tex]
[tex]26.2*10^-^4 hz[/tex] is frequency of the photon emitted by a hydrogen atom.
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what is the energy of a photon with a frequency of 8.34 x 10^14 hz, in joules?
Answer:
E = 5.52 x 10⁻¹⁹ J
Explanation:
The formula for energy of a photon, E, is:
[tex]E = hf[/tex]
where h is Plank's constant = 6.62 x 10 ⁻³⁴ .
Using the formula:
E = 6.62 x 10⁻³⁴ x 8.34 x 10¹⁴
= 5.52 x 10⁻¹⁹ J
The resistivity of pure water is fairly high, 1.8 × 105 Ω·m, whereas the resistivity of sea water is a million times lower, 2 × 10-1 Ω·m. Why does the high salt concentration make sea water significantly less resistive (i.e. more conductive) than pure water?
The high salt concentration make sea water significantly less resistive than pure water due to presence of charged ions in the sea water.
What is resistivity?The resistivity of a substance is the opposition to the flow of charges offered by the substance.
The greater the resistivity of a substance, the lesser its conductivity.
A high salt concentration make sea water highly conductive due to presence of charged ions in the water. The greater conductivity reduces the resistivity of the sea water.
A pure water has no charged ions in the water, thereby decreasing its conductivity and increasing its resistivity.
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A 3,204 kg tree positioned on the edge of a cliff 247 m above the ground breaks away and falls into the valley below which is considered zero potential energy. If the tree’s mechanical energy is conserved, what is the speed of the tree just before it hits the ground in meters/sec?
Let's see
PE is turned to KE as per law of conservation of energy[tex]\\ \rm\Rrightarrow mgh=\dfrac{1}{2}mv^2[/tex]
[tex]\\ \rm\Rrightarrow 2gh=v^2[/tex]
[tex]\\ \rm\Rrightarrow 2(10)(247)=v^2[/tex]
[tex]\\ \rm\Rrightarrow v²=4940[/tex]
[tex]\\ \rm\Rrightarrow v=70.3ms^{-1}[/tex]
