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3 years ago

The spectrum of light emitted by helium atoms is different than that emitted by hydrogen atoms because

Physics

1 answer:

MaRussiya [10]3 years ago

4 0

<h3><u>Answer;</u></h3>

The different atoms have different quantized energy levels

<h3><u>Explanation;</u></h3>

The atoms of different elements have different energy levels because they have different nuclear charges and spins, and different numbers of electrons.
Each different kind of atom, like hydrogen or radon, has a distinct nuclear charge and number of electrons. This makes the potential energy function different for each atom, and therefore results in different energy levels.
In an emmission spectra, each bright band corresponds to a difference between energy levels within the atom.

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Please HELP ASAP PLEASE I WILL MARK BRAINLIEST

Anika [276]

It’s True the volume is 1k of iron is = 1

3 0

3 years ago

Plz help ASAP I'll mark as brainliest

gogolik [260]

Hi there!

Hooke's law states that:

F = -kx

k = Spring constant (N/m)

x = DISPLACEMENT from equilibrium (m)

Essentially, the force of a spring is PROPORTIONAL to its spring constant and its displacement from its equilibrium point.

The force of the spring (T) is not proportional to the spring's length (l), but rather its DISPLACEMENT from its equilibrium length. (Δl)

The equilibrium length is where the force of the spring (T) = 0N. Looking at the graph, the line intersects this value at l = 30cm.

We can begin by looking at the given graph.

When the spring force = 4N, the total length of the spring is 35 cm.

Now, the EQUILIBRIUM length is 30 cm, so the total elongation is:

35 - 30 = 5 cm.

5.1.

If the spring elongates by 10 cm, the total length of the spring is:

30 + 10 = 40 cm

According to the graph, a length of 40 cm corresponds to a force of 8N.

5.2.

We can solve for the weight of the ball using the following:

W (weight) = m (mass) · acceleration due to gravity (10N/kg)

Using a summation of forces:

∑F = T - W

The elongation that we are solving for occurs at the equilibrium point (net force = 0 N), so:

0 = T - W

T = W = 8 N

5.3.

0 = T - Mg

T = Mg

Use the prior value of T and gravity to solve:

8 = 10M

m = 0.8 kg

8 0

2 years ago

What will be the average velocity of a body falling in free fall on Earth for 3 s?

SpyIntel [72]

Answer:

29.4m/s

Explanation:

Given parameters:

Time = 3s

Unknown:

Average velocity = ?

Solution:

To solve this problem, we use the expression below:

v = u + gt

v is the average velocity

u is the initial velocity = 0m/s

g is the acceleration due to gravity = 9.8m/s²

t is the time

So;

v = 0 + (9.8 x 3) = 29.4m/s

6 0

2 years ago

While passing through a human cell, an x-ray photon interacts with and inactivates the cell’s master molecule. What is the conse

Yuliya22 [10]

Answer:

Cell Death

Explanation:

Cell death is defined as the biological process which ceases the function of the cell to carry out. This can be caused due to the formation of new cells in place of old cells.

Or it can be cause due to some serious disease or may be caused due to the injury or due to the death of that organism to which these cells belong.

And another case is that when X-ray photon interact with the human cell while it passes through the cell, it will damage the cell and cease it to function well and a more drastic condition occurs and that cell become dead.

3 0

2 years ago

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

3 years ago