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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<h2>Let us predict the gas and liquid products in option </h2>
Explanation:
option 1 : 2 upper H g upper O (s) right arrow 2 upper H g (l) plus upper O subscript 2 (g).
Chemical reaction
It is the reactants react to form products .
Chemical equation
It is the method of representing reaction in terms of moles , specifying states , symbols , molecular formulas etc .
This actually gives the complete info about the reaction .
In the above asked question or any other question we can specify the states by writting :
Solid as "S".
Liquid as "L"
Gas as (g).
Answer:
The answer to your question is P = 1.357 atm
Explanation:
Data
Volume = 22.4 L
1 mol
temperature = 100°C
a = 0.211 L² atm
b = 0.0171 L/mol
R = 0.082 atmL/mol°K
Convert temperature to °K
Temperature = 100 + 273
= 373°K
Formula

Substitution

Simplify
(P + 0.0094)(22.3829) = 30.586
Solve for P
P + 0.0094 = 
P + 0.0094 = 1.366
P = 1.336 - 0.0094
P = 1.357 atm