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DiKsa [7]
2 years ago
14

Large and small numbers are often best entered in exponential notation. For example, the mass of the earth is 5.98x1024 kg and t

he mass of an electron is 9.11x10-31 kg. Canvas does not have the capability to handle scientific notation. We will include the power of 10 with the units needed in the answer. The charge of an electron is -1.60x10-19 C. Determine the charge to mass ratio of the electron (in 1011 C/kg). (Hint: divide the electron's charge by its mass; the answer is a negative number with a positive exponent)
Physics
2 answers:
stich3 [128]2 years ago
5 0

Answer:

The charge to mass ratio is

m/e= -0.17*10^12kg/c

Explanation:

Step one :

Given data

mass of an electron m= 9.11x10-31kg

charge of electron q= -1.60x10-19C

Applying the equation

mv²/r=Bev

Where m =mass of charge

v= velocity of charge

r= distance

B=magnetic field

e= charge

Rearranging to get the charge-mass ratio we have

mv²/r=Bev

m/e=Br/v

(-1.60x10-19)/(9.11x10-31kg)=Br/v

-0.17*10^12

m/e= -0.17*10^12kg/c

Doss [256]2 years ago
4 0

Answer:

The charge to mass ratio is -1.76\times10^{11}

Explanation:

Mass\ of\ electron=m_e=9.11\times10^{-31} kg\\ Charge\ of\ the\ electron=q_e=-1.60\times10^{-19} C\\

We need to find how much charge is contained  in the electron per unit of mass, to do this we divide the charge in an electron and the mass of an electron:

\frac{Charge\ of\ electron}{Mass\ of\ electron}=\frac{q_e}{m_e}\frac{C}{kg}=\frac{-1.60\times10^{-19} C}{9.11\times10^{-31} kg}=-1.76\times10^{11} \frac{C}{kg}

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The discovery and characterization of cathode rays was important in the development of the atomic theory because
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Answer:

All matter contained electrons

Explanation:

The discovery and characterization of cathode ray suggested that it was a subatomic particle and cathode ray ( electron) was the first discovered. It immensely became the strong explanatory tool for chemical bond. This can be attributed to  the the ease with which electron move from one atom to the other.  

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2 years ago
Which one of the following represents an acceptable set of quantum numbers for an electron in an atom? (arranged as n, l, m l ,
Vitek1552 [10]

Answer:

The correct option that represents an acceptable set of quantum numbers for an electron in an atom is;

(b) 4, 3, -3, 1/2.

Explanation:

To solve the question, we note that the available options where the set of quantum numbers for an electron in an atom are arranged as n, l, m l , and ms are;

4, 4, 4, 1/2

4, 3, -3, 1/2

4, 3, 0, 0

4, 5, 7, -1/2

4, 4, -5, 1/2

Let us label them as a to as follows

(a) 4, 4, 4, 1/2

(b) 4, 3, -3, 1/2

(c) 4, 3, 0, 0

(d) 4, 5, 7, -1/2

(e) 4, 4, -5, 1/2

Next we note the rules for the assignment and arrangement of quantum numbers are as follows

Number                                   Symbol                Possible values

Principal Quantum Number  .......n........................1, 2, 3, ......n

Angular momentum quantum

number...............................................l.........................0, 1, 2, .......(n - 1)

Magnetic Quantum Number........m₁......................-l, ..., -1, 0, 1,.....,l  

Spin Quantum Number.................m_s.....................+1/2, -1/2

We are meant to analyze each of the arrangement for acceptability.

Therefore for (a),

we note that the angular momentum quantum number, l =4 , is equal to the principal quantum number n =4 which violates the rule as the maximum value of the angular momentum quantum number is (n-1) where the maximum value of the principal quantum number is n.

Therefore (a) is not acceptable.

(b) Here we note that

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = -3 ∈ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (b) 4, 3, -3, 1/2 represents an acceptable set of quantum numbers for an electron in an atom.

(c) Here we have

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = 0 ∈ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 0 ∉ (+1/2, -1/2) → not acceptable

Therefore (c) 4, 3, 0, 0 does not represents an acceptable set of quantum numbers for an electron in an atom.

(d) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 5 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = 7 ∉ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = -1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (d) 4, 5, 7, -1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

(e) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 4 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = -5 ∉ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (e) 4, 4, -5, 1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

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A pendulum is used in a large clock. The pendulum has a mass of 2kg. If the pendulum is moving at a speed of 2.9 m/s when it rea
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Answer:

Explained

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2 years ago
jesse is swinging miguel in a circle at a tangential speed of 3.50 m/s. if the radius of the circle is 0.600 m and miguel has a
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That's the tension in Miguel's arm or leg or whatever part of his body
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2 years ago
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