All categories

1 year ago

How many electrons are located in the outermost orbit in the bohr model of a boron atom?.

Physics

1 answer:

Genrish500 [490]1 year ago

5 0

Electrons that are located in the outermost orbit in the Bohr model of a boron atom will be 3.

A nucleus with six neutrons and five protons makes up the Bohr model of boron. The K-shell and L-shell electron shells around the nucleus The valence electrons, which make up three of the electrons in the outermost shell of the Bohr diagram, are also present. The orbits of the electrons around the small nucleus are described by the Bohr model of boron.

It utilizes a number of different electron shells, including K, L, M, and N. The number of electrons that each shell can hold varies, with the shell closest to the nucleus having the least energy. The energy levels increase as the shell recedes more. According to the Bohr model, the electrons in an atom may be seen to orbit the nucleus in a variety of orbits with differing energies.

Learn to know more about Electronic configuration on

brainly.com/question/26084288

#SPJ4

You might be interested in

As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length

grin007 [14]

Answer:

As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your window is Lw. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g.

Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).

Question:

If the bottom of your window is a height hb above the ground, what is the velocity vground of the pot as it hits the ground? You may introduce the new variable vb, the speed at the bottom of the window, defined by

vb=Lwt+gt2.

Express your answer in terms of some or all of the variables hb, Lw, t, vb, and g.

The correct answer is

vground = (t²+lw²+(2t³+2t²)×g×Lw+(t⁴+2t³+t²) ) ∧ 0.5 = vb² +2g×vbt+1/2gt²

Explanation:

We have from the relation

v = u + gt, S = ut + 1/2 gt², v² = u² + 2gS

in this case S = hb, u = vb=Lwt+gt2. and v = vground

therefore v² = (Lwt+gt²)² + 2 × g × hb

= (Lwt+gt²)² + 2 × g × (Lwt+gt²)×t + 1/2 gt² = vb² +2g×vbt+1/2gt²

vground = (t²+lw²+(2t³+2t²)×g×Lw+(t⁴+2t³+t²) ) ∧ 0.5

7 0

3 years ago

A physical quantity X is connected from X = ab2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3

Hoochie [10]

Answer: 11%

Explanation:

Given that

X = ab^2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3 respectively.

Percentage error of b = 2%

Percentage error of b^2 = 2 × 2 = 4

When you are calculating for percentage error that involves multiplication and division, you will always add up the percentage error values.

Percentage error of X will be;

Percentage error of a + percentage error of b^2 + percentage error of c

Substitute for all these values

4 + 4 + 3 = 11%

Therefore, percentage error of X is 11%

3 0

3 years ago

Two blocks of clay, one of mass 100 kg and one of mass 3.00 kg, undergo a completely inelastic collision. Before the collision o

adoni [48]

Answer:

a) v = 0.8 m / s , b) $v_{f}$ = 0.777 m / s , c) ΔK = 0.93 J

Explanation:

This exercise can be solved using the concepts of moment, first let's define the system as formed by the two blocks, so that the forces during the crash have been internal and the moment is conserved.

They give us the mass of block 1 (m1 = 100kg, its kinetic energy (K = 32 J), the mass of block 2 (m2 = 3.00 kg) and that it is at rest (v₀₂ = 0)

Before crash

po = m1 vo1 + m2 vo2

po = m1 vo1

After the crash

$p_{f}$ = (m1 + m2) $v_{f}$

a) The initial speed of the block of m1 = 100 kg, let's use the kinetic energy

K = ½ m v²

v = √2K / m

v = √ (2 32/100)

v = 0.8 m / s

b) The final speed,

p₀ = $p_{f}$

m1 v₀1 = (m1 + m2) $v_{f}$

$v_{f}$ = m1 / (m1 + m2) v₀₁

The initial velocity is calculated in the previous part v₀₁ = v = 0.8 m / s

$v_{f}$ = 100 / (3 + 100) 0.8

$v_{f}$ = 0.777 m / s

c) The change in kinetic energy

Initial K₀ = $K_{f}$

K₀ = 32 J

Final $K_{f}$ = ½ (m1 + m2) $v_{f}$ ²

$K_{f}$ = ½ (3 + 100) 0.777²

$K_{f}$ = 31.07 J

ΔK = $K_{f}$ - K₀

ΔK = 31.07 - 32

ΔK = -0.93 J

As it is a variation it could be given in absolute value

Part D

For this part s has the same initial kinetic energy K = 32 J, but it is block 2 (m2 = 3.00kg) in which it moves

d) we use kinetic energy

v = √ 2K / m2

v = √ (2 32/3)

v = 4.62 m / s

e) the final speed

v₀₂ = v = 4.62 m/s

p₀ = m2 v₀₂

m2 v₀₂ = (m1 + m2) $v_{f}$

$v_{f}$ = m2 / (m1 + m2) v₀₂

$v_{f}$ = 3 / (100 + 3) 4.62

$v_{f}$ = 0.135 m / s

f) variation of kinetic energy

$K_{f}$ = ½ (m1 + m2) $v_{f}$ ²

$K_{f}$ = ½ (3 + 100) 0.135²

$K_{f}$ = 0.9286 J

ΔK = 0.9286-32

ΔK = 31.06 J

4 0

3 years ago

A test charge is introduced into the electric field of a charge. It feels a force of 2F where the electric field is 2E. What wou

dezoksy [38]

Answer:8F

Explanation:

8 0

4 years ago

A wheel of radius 30 cm is rotating at a rate of 2.0 revolutions every 0.080 s. (A) through what angle, in radians, does the whe

Alchen [17]

A)
2revs in 0.08s
so in 1s thats 25revs
therefore thats 50π radians in one second

b)
well, ω=2π/T
therefore ω=50π = 157.079rads^-1
and v=rω where r is in meters;
v=0.3x157.079
v=47.123ms^-1

c)
f=1/T
f=1/period for one rotation
1 rotation = 0.08/2 = 0.04
f=1/0.04
f=25Hz

6 0

3 years ago