All categories

4 years ago

How does the nuclear charge relate to the energy of the 1s orbital?

Physics

1 answer:

nlexa [21]4 years ago

3 0

Explanation:

Effective nuclear charge is defined as he net positive charge experienced by an electron in an atom. It is termed "effective" because the shielding effect of electrons prevents higher orbital electrons from experiencing the full nuclear charge of the nucleus due to the repelling effect of inner-layer electrons.

The 1s is the closest shell to the nucleus of an therefore maximum nuclear charge is experienced. The formula for effective nuclear charge is:

Zeff = Z – S

where

Z = the number of protons in the nucleus, and

S = the shielding constant, the average number of electrons between the nucleus and the electron.

Hence, the energy required to remove an electron from the 1s orbital is the strongest.

You might be interested in

A bicyclist starts from rest and accelerates at a rate of 2.3 m/s^2 until it reaches a speed of 23 m/s. It then slows down at a

Rama09 [41]

Answer:

33 seconds.

Explanation:

The equation for speed with constant acceleration at time t its:

$V(t) \ = \ V_0 \ + \ a \ t$

where $V_0$ is the initial speed, and a its the acceleration.

<h3>First half of the problem</h3>

Starting at rest, the initial speed will be zero, so

$V_0 = 0$

the final speed is

$V(t_{f1}) = 23 \frac{m}{s}$

and the acceleration is

$a = 2.3 \frac{m}{s^2}$ .

Taking all this together, we got

$V(t_{f1}) = 23 \frac{m}{s} = 0 + 2.3 \ \frac{m}{s^2} t_{f1}$

$23 \frac{m}{s} = 2.3 \ \frac{m}{s^2} t_{f1}$

$\frac{23 \frac{m}{s}}{2.3 \ \frac{m}{s^2}} = t_{f1}$

$10 s = t_{f1}$

So, for the first half of the problem we got a time of 10 seconds.

<h3>Second half of the problem</h3>

Now, the initial speed will be

$V_0 = 23 \frac{m}{s}$ ,

the acceleration

$a=-1.0 \frac{m}{s^2}$ ,

with a minus sign cause its slowing down, the final speed will be

$V(t_{f2}) = 0$

Taking all together:

$V(t_{f2}) = 0 = 23 \frac{m}{s} - 1.0 \frac{m}{s^2} t_{f2}$

$23 \frac{m}{s} = 1.0 \frac{m}{s^2} t_{f2}$

$\frac{23 \frac{m}{s}}{1.0 \frac{m}{s^2}} = t_{f2}$

$23 s = t_{f2}$

So, for the first half of the problem we got a time of 23 seconds.

<h3>Total time</h3>

$t_total = t_{f1} + t_{f2} = 33 \ s$

5 0

3 years ago

How do I calculate the value of the slope graph

marysya [2.9K]

Answer:

Slope is calculated using the equation: slope= rise/run (y/x). Take take y value of a point and put it over the x value of the same point and then simplify the fraction.

Explanation:

8 0

3 years ago

Physical activity increases your risk for coronary artery disease. true false

goldfiish [28.3K]

Answer:

false here is why :)

Explanation:

Regular, moderate-intensity physical activity reduces the risk of developing coronary heart disease and can reduce the likelihood of dying from heart disease.

hope this helped.

7 0

3 years ago

If the same change in velocity occurs in less time, does the magnitude of the corresponding average acceleration increase, decre

Zielflug [23.3K]

Answer:

increase

Explanation:

As acceleration is the time rate of change of velocity

a = Δv/t

with t in the denominator, smaller time values result in greater acceleration values with constant velocity change.

7 0

3 years ago

Suppose you first walk 12.5 m in a direction 20° west of north and then 24 m in a direction 40° south of west as shown in the fi

Firlakuza [10]

La ce materie vr ca sunt pe tableta

3 0

3 years ago