Of the following transitions in the bohr hydrogen atom, the transition results in the emission of the highest-energy photon.

a. n=6 to n=1
b. n=1 to n=6
c. n=1 to n=5
d. n=5 to n=1

n=6 to n=1 transition results in the emission of the highest-energy photon.



Emission of protons often goes from a higher energy level down to a lower energy level.Bohr shows the hydrogen atom looking at an electron, often moves in a circular orbit in a nucleus. These transitions in the orbits brings about in the absorption or emission of photons. When an electron moves from a higher-energy orbit to a more stable one, energy is given off in the form of a photon.

Conclusively we can therefore say that the answer is option A because the n=1 energy level is at the ground state energy level, which therefore implies that energy is the greatest in this energy level

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Option A. n = 6 to n = 1

Explanation:

Niels Bohr was a Danish physicist who proposed the hydrogen atom quantum model to explain the discontinuity of the atom's emission spectra.

In Bohr hydrogen atom model, the electrons occupy orbits identified with the numbers n = 1, 2, 3, 4, ... Each number (orbit) corresponds to a different energy level  or state. The number n = 1 corresponds to the lowest energy level, and each higher number corresponds to a higher energy level.

This table shows the relative energy of the different orbits of the Bhor hydrogen atom:

Orbit      Quantum       Energy      Relative

              number         level          energy

First          n = 1                  1                   E₁

Second    n = 2                 2                2E₁

Third        n = 3                 3                 9E₁

Fourth      n = 4                 4                16E₁

Fifth          n = 5                5                25E₁

Sixth         n = 6                6                36E₁

Seventh    n = 7                7                49E₁

When an electron jumps from a higher energy state down to a lower energy state, it emits a photon with an energy equal to the difference of the energies between the initial and the final states.

Since the n = 6 to n = 1 transition results in the higher relative energy difference (36E₁ - E₁ = 35E₁), you conclude that it is this transition which results in the emission of the highest-energy photon, which is the option A.

The answer is false because: gravitational force is independent of medium and depends only on the masses of the two objects and the distance between them

Answer is:     molar solubility of  yttrium fluorid is 4.23·10⁻⁶ m. balanced chemical reaction: yf₃(s)  → y³⁺(aq) + 3f⁻(aq). ksp = 8.62·10⁻²¹. ksp = [y³⁺]  · [f⁻]³.[y³⁺] = [yf₃] = s.[f⁻] = 3[yf₃] = 3s.ksp(yf₃)  = s  · (3s)³.8.62·10⁻²¹ = 27s⁴.s(yf₃)  =  ⁴√(8.62·10⁻²¹  ÷ 27).s(yf₃)  = 0.00000423 m.s(yf₃) = 4.23·10⁻⁶ m.