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2 years ago

Which of electromagnetic radiation has the shortest wavelength?

Physics

1 answer:

9966 [12]2 years ago

4 0

Answer:

The electromagnetic radiation with the shortest wavelength are gamma rays.

Explanation:

Photons are the particles that constitute light and its kwnow that lights propagates as electromagnetic waves.

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensities, that radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

The distribution of the radiation in the electromagnetic spectrum can also be given in wavelengths, but it is more frequent to work with it at frequencies, the highest being that of gamma rays, followed by X-rays, ultraviolet rays and the visible region , and those of lower frequencies, which correspond to infrared, microwave and radio waves.

The energy given for each wavelength can be found by means of the following equation.

$E = h\nu$ (1)

but $c = \lambda \cdot \nu$

$\nu = \frac{c}{\lambda}$

$E = \frac{hc}{\lambda}$ (2)

Where E is the energy, h is the planck's constant and $\lambda$ is the wavelength.

So, it can be see in equation 2 that when the wavelength increases the energy decreases (inversely proportional).

Hence, the electromagnetic radiation with the shortest wavelength are gamma rays.

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In the Bohr model of the atom, atomic electrons approximatately 'orbit' the nucleus. The hydrogen atom consists of a proton of m

lara31 [8.8K]

Answer:

$F=3.61\times 10^{-47}\ N$

Explanation:

Mass of a proton, $m_p=1.67\times 10^{-27}\ kg$

Mass of an electron, $m_e=9.11\times 10^{-31}\ kg$

The distance between the electron and the proton is, $r=5.3\times 10^{-11}\ m$

We need to find the mutual attractive gravitational force between the electron and proton. The gravitational force is given by :

$F=G\dfrac{m_em_p}{r^2}$

Where G is the universal Gravitational constant

$F=6.67\times 10^{-11}\times \dfrac{9.11\times 10^{-31}\times 1.67\times 10^{-27}}{(5.3\times 10^{-11})^2}\\\\F=3.61\times 10^{-47}\ N$

So, the force between the electron and proton is $3.61\times 10^{-47}\ N$ .

7 0

3 years ago

Based on its type of chemical bond, which of the following has the highest boiling point?

Sever21 [200]

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

Potassium chloride

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

Potassium Chloride is an ionic compound and has ionic bond which is stronger than covalent bond in ethyl alcohol,water ,ammonia, and thus has the highest boiling point .
ionic bond is a type of bond that results from the transfer of electrons between metallic atoms and non-metallic atoms.

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2 years ago

two thermometers, calibrated in celsius and fahrenheit respectively, are put into a liquid. the reading on the fahrenheit scale

Viefleur [7K]

Two thermometers, calibrated in celsius and fahrenheit respectively, are put into a liquid. the reading on the fahrenheit scale is four times the reading on the celsius scale. the temperature of the liquid is:

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3 years ago

A projectile falls beneath the straight-line path it would follow if there were no gravity. How many meters does it fall below t

maria [59]

Answer:

Explanation:

Suppose v is the initial velocity and $\theta$ is the angle of inclination

distance traveled in vertical direction in t=1 s

When gravity is present

$y=vt+\frac{1}{2}at^2$

where $y=vertical\ distance$

$a=acceleration$

$t=time$

$v=initial\ velocity$

here initial velocity is v\sin \theta [/tex] so

$y=v\sin \theta \times 1-\frac{1}{2}gt^2$

$y=v\sin \theta -0.5g$

In absence of gravity

$y_2=v\sin \theta \times t$

$y_2=v\sin \theta \times 1$

$\Delta y=y_2-y=v\sin \theta -v\sin \theta +0.5 g=4.9\ m$

4 0

3 years ago

The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the

Eddi Din [679]

Answer:

x = D (M/M-m) 2.41

Explanation:

a) Let's apply Newton's second law to find the summation of force, where each force is given by the law of universal gravitation

F = g m₁m₂ / r²

Σ F = 0

F1- F2 = 0

F1 = F2

We set the reference system in the body of greatest mass (M) the planet

F1 = g m₁ M / x²

F2 = G m1 m / (D-x)²

G m₁ M / x² = G m₁ m / (D-x)²

M (D-x)² = m x²

MD² -2MD x + M x² = m x²

x² (M-m) -2MD x + MD² = 0

We solve the second degree equation

x = [2MD ±√ (4M²D² - 4 (M-m) MD²)] / 2 (M-m)

x = {2MD ± 2D √ (M² + (M-m) M)} / 2 (M-m)

x = D {M ± Ra (2M²-mM)} / (M-m)

x = D (M ± M √ (2-m/M)) / (M-m)

x = D (M / (M-m)) (1 ±√ (2-m/M)

Let's analyze this result, the value of M-m >> 1, so if we take the negative root, the value of x would be negative, it is out of the point between the two bodies, so the correct result must be taken with the positive root

x = D (M / (M-m)) (1 + √2)

x = D (M/M-m) 2.41

b) X = 2/3 D

x = D (M/M-m) 2.41

2/3 D = D (M/(M-m)) 2.41

2/3 (M-m) = M 2.41

2/3 M - 2/3 m = 2.41 M

1.743 M = 0.667 m

M/m = 0.667/1.743

M/m = 0.38

3 0

2 years ago