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gizmo_the_mogwai [7]
3 years ago
6

(a) What is the ionization energy of a hydrogen atom that is in the n = 6 excited state? (b) For a hydrogen atom, determine the

ratio of the ionization energy for the n = 6 excited state to the ionization energy for the ground state.
Physics
1 answer:
crimeas [40]3 years ago
7 0

Answer:

(a) 0.3778 eV

(b) Ratio = 0.0278

Explanation:

The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

E=\frac {-13.6}{n^2}\ eV

(a) The energy of the electron in n= 6 excited state is:

E=\frac {-13.6}{6^2}\ eV

E=-0.3778\ eV

Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV

(b) For first orbit energy is:

E=\frac {-13.6}{1^2}\ eV

E=-13.6\ eV

Ratio=\frac {E_6}{E_1}

Ratio=\frac {-0.3778}{-13.6}

Ratio = 0.0278

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Answer:

Energy of scattered photon is 232.27 keV.

Kinetic energy of recoil electron is 497.73 keV.

The recoil angle of electron is 13.40°

Explanation:

The energy of scattered photon is given by the relation :

E_{2}=\frac{E_{1} }{1+(\frac{E_{1} }{m_{e}c^{2}  })(1-\cos\theta) }     .....(1)

Here E₁ is the energy of incident photon, E₂ is the energy of scattered photon, m_{e} is mass of electron and θ is scattered angle.

Substitute 730 keV for E₁, 511 keV for m_{e} and 120° for θ in equation (1).  

E_{2}=\frac{730 }{1+(\frac{730 }{511  })(1-\cos120) }

E₂ = 232.27 keV

Kinetic energy of recoil electron is given by the relation :

K.E. = E₁ - E₂ = (730 - 232.27 ) keV = 497.73 keV

The recoil angle of electron is given by :

\cot\phi=(1+\frac{E_{1} }{m_{e}c^{2}  })\tan\frac{\theta}{2}

Substitute the suitable values in above equation.

\cot\phi=(1+\frac{730 }{511  })\tan\frac{120}{2}

\cot\phi=4.20

\phi = 13.40°

8 0
3 years ago
Which set of electromagnetic waves is in the correct order from greatest to least energy?
kifflom [539]

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Explanation:

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If a girl walks 6km due west changes directions and walk another 5km due north. Find her displacement in both magnitude and dire
xenn [34]

Answer:

the girl's displacement in both magnitude and direction is 7.81 m at 50.2⁰ North west.

Explanation:

Given;

6km due west, and

5km due north

The magnitude of her displacement is calculated by forming a right angled triangle. The hypotenuse side of the triangle is the girl's displacement.

d² = 5² + 6²

d² = 25 + 36

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d = √61

d = 7.81 m

The direction of the girl is calculated as;

tan \ \theta = \frac{6}{5} \\\\tan \ \theta = 1.2\\\\\theta = tan^{-1} (1.2)\\\\\theta = 50.2^0

Therefore, the girl's displacement in both magnitude and direction is 7.81 m at 50.2⁰ North west.

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A 2kg body is moving a long a horizontal circular path of radius 2m with constant speed of 6m/s . what is the magnitude of the f
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Each of the rods depicted below were machined from same stock metal. They were originally machined to be the same length, but th
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The force required to extend a rod increases as the cross sectional area

increases.

The rod that experiences the largest force is <u>rod B</u>

Reason:

The elongation of a rod by the application of a force is given by the

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\Delta L = \dfrac{F \cdot L}{A \cdot E}

From the above equation, we have that the elongation is inversely

proportional to the cross sectional area, such that the extension of a rod by

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Therefore, the force required to extend the length of a rod by a specific

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indicating that the rod with the largest cross sectional area require the

most force and therefore, experiences the largest force.

The rod that experiences the largest force is the rod with the largest cross

sectional area, which is <u>rod B</u>

Learn more here:

brainly.com/question/12937199

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2 years ago
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