The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>
the diagram of the emission spectrum has been added.
<em>From the given</em><em> chart;</em>
The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m
The frequency of this emission is calculated as follows;
c = fλ
where;
- <em>c is the speed of light = 3 x 10⁸ m/s</em>
- <em>f is the frequency of the wave</em>
- <em>λ is the wavelength</em>

The energy of the emitted photon corresponding to the orange line is calculated as follows;
E = hf
where;
- <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>
<em />
E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)
E = 3.26 x 10⁻¹⁹ J.
Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
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Answer:
The correct option is B
Explanation:
One of the claims of John Dalton's atomic theory is that atom is the smallest unit of matter (which suggests that there are no particles smaller than an atom in any matter). This claim has been disproved by the modern atomic theory which established that there are particles smaller than atom (called subatomic particles). These particles are electrons, protons and neutrons.
One of the modern atomic theory was by Neils Bohr, who proposed that <u>electrons move in circular orbits around the central nucleus</u>. Thus, the electrons of iron can also be said to be present in a region of space (circular path) around the nucleus. This proves that option B is the correct option as John Dalton's theory did not even recognize the electron(s) nor the nucleus.
This question requires the knowledge of density.
The density of ethyl alcohol = 789 kg m⁻³
The density of water = 1000 kg m⁻³
Density = Mass / Volume
By applying ethyl alcohol,
789 kg m⁻³ = Mass / 0.9 m³
Mass = 710.1 kg
hence the mass of 0.9 m³ ethyl alcohol is 710.1 kg.
Then by applying water,
1000 kg m⁻³ = 710.1 kg / Volume
Volume = 0.7101 m³
= 0.7 m³
hence the equal water volume is 0.7 m³
Answer:
The volume of an irregularly shaped solid can be determined from the volume of water it displaces. A graduated cylinder contains 19.9 mL of water. When a small piece of galena is added, it sinks and the volume increases to 24.5 mL
Answer:
If 1,079.75 Joules of heat are added to 77.75 grams of water, by 3.32 degrees Celsius the temperature of water will increase

Explanation:

Here , q = heat added / removed from the substance
m = mass of the substance taken
= Change in temperature
C = specific heat capacity of the substance
In liquid state the value of C for water is :

Given values :
q = 1079.75 J
m = 77.75 gram
Insert the value of C, m , q in the given equation
on transposing ,


