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3 years ago

What is the energy level of hydrogen?

Physics

1 answer:

barxatty [35]3 years ago

8 0

Answer:

The energies corresponding to each of the allowed orbitals are called energy levels.

Explanation:

A scientist known as Niels Bohr put forward that electrons in an atom covers some permitted orbitals with a specific energy. In other words, the energy of an electron in an atom is not continuous, but 'quantized.' The energies corresponding to each of the allowed orbitals are called energy levels.

$E = -\frac{E_0}{n^2} \\where \\E_0 = 13.6 eV (1 eV = 1.602\times 10^{-19}Joules)\\and\ n = 1,2,3...$

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local AM radio station broadcasts at a frequency of 685.9 kHz. Calculate the wavelength at which they are broadcasting. Waveleng

natta225 [31]

Answer:

λ = 437 m.

Explanation:

Since a radio wave is a electromagnetic type of wave, it propagates approximately at the same speed of light in vacuum, 3*10⁸ m/s.
As in any wave, there exist a fixed relationship between the propagation speed, the wavelength and the frequency of the radio wave, as follows:

$c = \lambda* f (1)$

Replacing by the values of the speed and the frequency, and solving for the wavelength λ, we get:

$\lambda = \frac{c}{f} =\frac{3e8 m/s}{685.9e31/s} =437 m (2)$

8 0

3 years ago

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting a

IRISSAK [1]

A. $4.64\cdot 10^{11}m$

The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:

$v=\frac{2\pi r}{T}$

where we have:

$v=30,000 km/s = 3\cdot 10^7 m/s$ is the orbital speed

r is the orbital radius

$T=27 h \cdot 3600 =97,200 s$ is the orbital period

Solving for r, we find the distance of the clumps of matter from the centre of the black hole:

$r=\frac{vT}{2\pi}=\frac{(3\cdot 10^7 m/s)(97200 s)}{2\pi}=4.64\cdot 10^{11}m$

B. $6.26\cdot 10^{36}kg, 3.13\cdot 10^6 M_s$

The gravitational force between the black hole and the clumps of matter provides the centripetal force that keeps the matter in circular motion:

$m\frac{v^2}{r}=\frac{GMm}{r^2}$

where

m is the mass of the clumps of matter

G is the gravitational constant

M is the mass of the black hole

Solving the formula for M, we find the mass of the black hole:

$M=\frac{v^2 r}{G}=\frac{(3\cdot 10^7 m/s)^2(4.64\cdot 10^{11} m)}{6.67\cdot 10^{-11}}=6.26\cdot 10^{36}kg$

and considering the value of the solar mass

$M_s = 2\cdot 10^{30}kg$

the mass of the black hole as a multiple of our sun's mass is

$M=\frac{6.26\cdot 10^{36} kg}{2\cdot 10^{30} kg}=3.13\cdot 10^6 M_s$

C. $9.28\cdot 10^9 m$

The radius of the event horizon is equal to the Schwarzschild radius of the black hole, which is given by

$R=\frac{2MG}{c^2}$

where M is the mass of the black hole and c is the speed of light.

Substituting numbers into the formula, we find

$R=\frac{6.26\cdot 10^{36} kg)(6.67\cdot 10^{-11})}{(3\cdot 10^8 m/s)^2}=9.28\cdot 10^9 m$

8 0

3 years ago

A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its veloci

Vlad [161]

Answer:

Approximately $\rm 2.5\; m \cdot s^{-1}$ .

Explanation:

Let the increase in the rocket's velocity be $\Delta v$ . Let $v_0$ represent the initial velocity of the rocket. Note that for this question, the exact value of $v_0$ doesn't really matter.

The momentum of an object is equal to its mass times its velocity.

Mass of the rocket with the 5 kg of fuel: $1000$ .
Initial velocity of the rocket and the fuel: $v_0$ .
Hence the initial momentum of the rocket: $1000\,v_0$ .

Mass of the rocket without that 5 kg of fuel: $1000 - 5 = 995$ .
Final velocity of the rocket: $v_0 + \Delta v$ .
Hence the final momentum of the rocket: $995\,(v_0 + \Delta v)$ .

Mass of the 5 kg of fuel: $5$ .
Final velocity of the fuel: $v_0 - 500$ (assuming that the the 500 m/s in the question takes the rocket as its reference.)
Hence the final momentum of the fuel: $5\,(v_0 - 500)$ .

Momentum is conserved in an isolated system like the rocket and its fuel. That is:

Sum of initial momentum = Sum of final momentum.

$1000\,v_0 = 995\,(v_0 + \Delta v) + 5\,(v_0 - 500)$ .

Note that $1000\, v_0$ appears on both sides of the equation. These two terms could hence be eliminated.

$0 = 995\, \Delta v - 5\times 500$ .

$\displaystyle \Delta v = \frac{5}{995}\times 500 \approx \rm 2.5\; m \cdot s^{-1}$ .

Hence, the velocity of the rocket increased by around 2.5 m/s.

5 0

3 years ago

You are running on a perfectly circular track that has a radius of 102 meters. What is the largest displacement you will ever at

tiny-mole [99]

The largest possible displacement on a circular track is the straight-line distance between the starting point and the point directly opposite it, half-way around the circle. That's the diameter of the track ... 204 meters.

8 0

3 years ago

Which of the following can be used to measure temperature accurately

Illusion [34]

Answer:

Any of those terms can be converted to either of the other terms, so either term is correct. People are accustomed to everyday temperatures in Fahrenheit. The ideal gas law specifies that

P V = N R T where T is in Kelvin which is Celsius + 273 deg.

4 0

2 years ago