The expiration date on a product refers to the last date a product should be _________

Fred has a very detailed model of the solar system where each planet is made out of granite rock. Since his little sister really

eimsori [14]

C or "Density" The granite rock has a much higher density than the Styrofoam.

3 0

3 years ago

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Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to

jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x $10^{-8}$ J.

(b) The wavelength of the photon is 1.2 x $10^{-17}$ m.

(c) The frequency of the photon is 2.47 x $10^{25}$ Hz.

Explanation:

Let;

$E_{1}$ = -13.60 ev

$E_{2}$ = -3.40 ev

(a) Energy of the emitted photon can be determined as;

$E_{2}$ - $E_{1}$ = -3.40 - (-13.60)

= -3.40 + 13.60

= 10.20 eV

= 10.20(1.6 x $10^{-9}$ )

$E_{2}$ - $E_{1}$ = 1.632 x $10^{-8}$ Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x $10^{-8}$ Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x $10^{-34}$ Js), c is the speed of light (3 x $10^{8}$ m/s) and λ is the wavelength.

10.20(1.6 x $10^{-9}$ ) = (6.6 x $10^{-34}$ * 3 x $10^{8}$ )/ λ

λ = $\frac{1.98*10^{-25} }{1.632*10^{-8} }$

= 1.213 x $10^{-17}$

Wavelength of the photon is 1.2 x $10^{-17}$ m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x $10^{-8}$ = 6.6 x $10^{-34}$ x f

f = $\frac{1.632*10^{-8} }{6.6*10^{-34} }$

= 2.47 x $10^{25}$ Hz

Frequency of the emitted photon is 2.47 x $10^{25}$ Hz.

6 0

3 years ago

The force of Earth's gravity on a capsule in space will lessen as it moves farther away. If the capsule moves to twice its dista

Bess [88]

Answer: One quarter of the force

Explanation:

According to Newton's law of Gravitation, the force $F$ exerted between two bodies of masses $m1$ and $m2$ and separated by a distance $r$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{(m1)(m2)}{r^2}$ (1)

Where $G$ is the gravitational constant

This means that the gravity force decreases when the distance between these two bodies increases.

In this context, if the distance between the capsule and the Earth increases twice, the new distance will be $2r$ .

Substituting this distance in (1):

$F=G\frac{(m1)(m2)}{(2r)^2}$ (2)

$F=G\frac{(m1)(m2)}{4r^2}$

<u>Finally:</u>

$F=\frac{1}{4}G\frac{(m1)(m2)}{r^2}$ >>>This means the force toward Earth becomes one quarter "weaker"

3 0

3 years ago

The ears are ____ and ____ to the shoulders and ____ to the nose.

Levart [38]

The ears are superior and posterior to the shoulders and lateral to the nose.

3 0

3 years ago

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An electron in the n = 7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 397 nm. what is the value

Dimas [21]

Answer:

$n_f=2$

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})$