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4 years ago

The energy required to remove an electron from a gaseous atom is called

Physics

1 answer:

IRISSAK [1]4 years ago

6 0

Answer: The energy required to remove an electron from a gaseous atom is called ionization energy.

Explanation:

Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.

When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period but decreases along a group.

Smaller is the size of an atom more will be the force of attraction between its protons and electrons. Hence, more amount of energy is required to remove an electron.

Thus, we can conclude that the energy required to remove an electron from a gaseous atom is called ionization energy.

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Find the minimum value of n in the Balmer series for which the predicted wavelength is in the ultraviolet region of the spectrum

d1i1m1o1n [39]

Answer:

λ =365.4 nm

Explanation:

Boh's atomic model of the Hydrogen atom the energy of each level is

Eₙ = - 13.606 / n²

where the synergy is in electonvotes and the value of E₀ = 13.606 eV is the energy of the base state of hydrogen.

An atomic transition occurs when an electron goes from an excited state and joins everything of lower energy.

ED = 13.606 (1 / n₀² - 1 / $n_{f}^{2}$ )

we are going to apply this relationship to answer slash.

At the beginning of the studies of atomic transitions, each group did not consider having a different name

name Initial state

Lymman 1

Balmer 2

the final state is any other state sta the continuum that corresponds to n = inf

Let's look for the highest energy of the Balmer series

ΔE = 13.606 (1/2² - 1 /∞)

ΔE = 3.4015 eV

Let's use the Planck relation for the energy

E = h f = h c /λ

λ = h c / E

Let's reduce the energy to J

E = 3.4015 eV (1.6 10⁻¹⁹ J / 1 eV) = 5.4424 10⁻¹⁹

λ = 6.63 10⁻³⁴ 3 10⁸ / 5.4424 10⁻¹⁹

λ = 3.654 10⁻⁷ m

λ = 3,654 10⁻⁷ m (10⁹ nm / 1m)

λ =365.4 nm

this eta radiation in the ultraviolet range

5 0

3 years ago

Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject

Finger [1]

Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

saturn orbit around the sun R_2 = 1.43 *10^9 km

earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

$t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}$

where

$\mu_{sun}$ is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2. $t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}$

t = 12,105.96 sec

6 0

4 years ago

Give example of not reversible change...

Gelneren [198K]

Answer:

There were many examples of this. One of the most well-known not reversible reaction is a precipitation reaction, in which an insoluble solid is formed from two aqueous solutions. An example is the reaction between silver nitrate and sodium chloride which forms a silver chloride precipitate.

3 0

3 years ago

A ball attached to a string is whirled around in a horizontal circle having a radius R. If the radius of the circle is changed t

ivanzaharov [21]

Answer:

twice the original speed

Explanation:

the centripetal force is the force of an object moving in a circle of turning and it is given by the expression

$F=\frac{mv^{2} }{r}$

where f = force of the object

m = mass

v = velocity

r = radius of the circle

first we need to know what the orignal speed will be then we can tell what will happen if the radius of the circle is changed to 4R.

from the expression for the centripetal force, we need to make the velocity subject of the formula

$F = \frac{mv^{2} }{r} \\cross multiplying\\Fr =mv^{2} \\dividing both sides by m\\\frac{Fr}{m}=v^{2} \\taking the square root of both sides\\v=\sqrt{\frac{Fr}{m} }$

now, if the radius is changed to 4R

in the expression for v above i will need to just insert 4R in place of r

$v=\sqrt{\frac{4rF}{m} } \\$

the square root of 4 is 2,

therefore

$v=2\sqrt{\frac{rF}{m} }$

we can see that this is just twice of what the original speed was.

8 0

4 years ago

Where does kinetic energy go when it is lost

kondor19780726 [428]

While the total energy of a system is always conserved, the kinetic energy carried by the moving objects is not always conserved. In an inelastic collision, energy is lost to the environment, transferred into other forms such as heat.

Hope this helps.

4 0

4 years ago