All categories

3 years ago

For each value of the principal quantum number n, what are the possible values of the electron spin quantum number m_s?

Physics

1 answer:

sergij07 [2.7K]3 years ago

8 0

Answer:

d) and e)

Explanation:

For any value of the principal quantum n, the only possible values of the electron spin quantum number m_s are + 1/2 and -1/2.
Since this number is related with the angular momentum of the electron, it has a magnitude (1/2) and an orientation given by the sign (+ or -)
So, the only right answers are d) and e).

You might be interested in

I need help on these 2 questions please ???

soldi70 [24.7K]

Friction plays a roll of operation in a car because we u stop pressing the peddle friction slowly starts making you slow down and stop.

3 0

4 years ago

Which of the following is most critical for our study of outter space

nordsb [41]

In my opinion, the answer would be d: a universal language. D would technically be more important because there are so many foreign scientific companies and to get the information we need to study the universe, we should communicate and give eachother what we have of our resources as we work for it.

6 0

3 years ago

Blood pressure is usually expressed by two numbers. What are those two numbers called?

sergiy2304 [10]

Top number is your systolic pressure

Bottom number is your diastolic pressure

Hoped this helped

8 0

3 years ago

Read 2 more answers

A 2.81 μF capacitor is charged to 1220 V and a 6.61 μF capacitor is charged to 560 V. These capacitors are then disconnected fro

fenix001 [56]

Answer:

756.88 Volts will be the potential difference across each capacitor.

Explanation:

$Q=C\times V$

Q = Charge on capacitor

C = Capacitance

V = Voltage across capacitor

Capacitance of first capacitor = $C_1=2.81 \mu F=2.81\times 10^{-6} F$

Charge of first capacitor = $Q_1$

Voltage across first capacitor = $V_1=1220 V$

$Q_1=C_1V_1$

$Q_1=2.81\times 10^{-6} F\times 1220 V=0.0034282 C$

Capacitance of first capacitor = $C_2=6.61\mu F=6.61\times 10^{-6} F$

Charge of second capacitor = $Q_2$

Voltage across first capacitor = $V_2=560 V$

$Q_2=C_2V_2$

$Q_1=6.61\times 10^{-6} F\times 560 V=0.0037016 C$

Both the capacitors are disconnected and positive plates are now connected to each other and the negative plates are connected to each other. These capacitors are connected in parallel combination.

Total charge = Q

$Q =Q_1+Q_2=0.0034282 C+ 0.0037016 C=0.0071298 C$

Total capacitance in parallel combination:

$C_p=C_1+C_2=2.81\times 10^{-6} F+6.61\times 10^{-6} F=9.42\times 10^{-6} F$

Potential across both capacitors = V

$V=\frac{Q}{C}=\frac{0.0071298 C}{9.42\times 10^{-6} F}=756.88 V$

756.88 Volts will be the potential difference across each capacitor.

8 0

3 years ago

when two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of

Ahat [919]

<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q</h2>

Explanation:

Specific heat capacity

It is defined as amount of heat required to raise the temperature of a substance by one degree celsius .

It is given as :

Heat absorbed = mass of substance x specific heat capacity x rise in temperature

or ,

Q= m x c x t

In above question , it is given :

For Q

mass of Q = m

Temperature changed =T₂/2

Heat supplied = x

Q= mc t

X=m x C₁ X T₁

or, X =m x C₁ x T₂/2

or, C₁=X x 2 /m x T₂ (equation 1 )

For another quantity : P

mass of P =m/2

Temperature= T₂

Heat supplied is same that is : X

so, X= m/2 x C₂ x T₂

or, C₂=2X/m. T₂ (equation 2 )

Now taking ratio of C₂ to c₁, We have

C₂/C₁= 2X /m.T₂ /2X /m.T₂

so, C₂/C₁= 1/1

so, the ratio is 1: 1

8 0

4 years ago