All categories

3 years ago

What is the difference between an atom in the ground state and an atom in an excited state

Physics

2 answers:

SIZIF [17.4K]3 years ago

7 0

The atom in an excited state has more energy and is less stable than the atom in the ground state.

alexgriva [62]3 years ago

6 0

Answer:

The ground state of an atom is defined as the lowest available energy the atom possesses.

An atom is said to be in an excited state when the atom gains energy and uses that energy to change its orbital arrangement.

Explanation:

You might be interested in

When the direction of acceleration is opposite to the direction of velocity

wariber [46]

Answer:

yes this is truly the wrong answer

7 0

3 years ago

Read 2 more answers

Question 1

galina1969 [7]

1/f= 1/di+1/do
1/f=1/90m+1/40m
1/f ≈27.7m

7 0

3 years ago

A tennis ball is dropped from a height of 3 m and bounces back to a height of

julsineya [31]

Answer:

To decide where the balls land, we need to determine how long the balls are in the air. Both balls will take 2 seconds to hit the ground.

Explanation:

1) Time played forward : gravity & drag forces are in opposite directions so it takes a longer time to reach the ground. 2) Time played backward : gravity & drag forces are in the same direction so it takes a shorter time to reach the ground.

5 0

2 years ago

The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

kotykmax [81]

Answer:

$2.1\times 10^{-12} c$

Explanation:

We are given that

Surface area of membrane= $5.3\times 10^{-9} m^2$

Thickness of membrane= $1.1\times 10^{-8} m$

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

$C=\frac{k\epsilon_0 A}{d}$

Substitute the values then we get

Capacitance between parallel plate capacitor= $\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}$

$C=0.25\times 10^{-12}F$

V= $85.9 mV=85.9\times 10^{-3}$

$Q=CV$

$Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c$

Hence, the charge resides on the outer surface= $2.1\times 10^{-12} c$

5 0

3 years ago

Read 2 more answers

I don’t understand how to do this

anzhelika [568]

Explanation:

A) Use Hooke's law to find the spring constant.

F = kx

40 N = k (0.4 m)

k = 100 N/m

B) Period of a spring-mass system is:

T = 2π √(m / k)

T = 2π √(2.6 kg / 100 N/m)

T = 1 s

Frequency is the inverse of period.

f = 1 / T

f = 1 Hz

5 0

3 years ago