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4 years ago

What is the relationship between wavelength of light and the quantity of energy per photon?

Physics

2 answers:

Andrew [12]4 years ago

8 0

Answer:

It is an inverse relationship

Explanation:

The energy of a photon is given by

e = hc/λ

Where e= energy of the photon, where c =velocity of light, h=Planck's constant and λ =wavelength of light

In this formula the energy is the numerator and the wavelength is the denominator. It can be written as

e= 1/λ .

This means for every increase in energy of photon there will be a decrease in wavelength and for every decrease in energy of the photon there will be an increase in the wavelength.

slamgirl [31]4 years ago

3 0

Answer:

The quantity of energy per photon is inversely proportional to the wavelength of the light.

Explanation:

Energy of light is given as

E = hf

where E = energy of the photons,

f = frequency of the light

If the number of photons = n

(E/n) = (h/n) f

Let (E/n) = E'

(h/n) = h'

But the frequency of light is related to wavelength through the relation

v = fλ

where v = speed of light = c

λ = wavelength of light

f = (c/λ)

E' = h' f

Substituting for f

E' = h' (c/λ)

h' and c are both constants, h'×c = K

E' = (K/λ)

So, the quantity of energy per photon is inversely proportional to the wavelength of the light.

Hope this Helps!!!

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A 75 A resistor in a circuit has a current flowing through it of 2.0 A. What is

Reptile [31]

Answer:

The power dissipated by the resistor can be calculated as $P = I^{2} R$ .

Explanation:

Joules heating law states that any conductor passed with the electric current produces heating effect which is the electric power loss.

The given circuit has, $R = 75 ohm$ resistor (the unit is wrong i question because resistor's unit is ohm not Ampere).
Current flowing through the circuit is $I = 2.0 A$ .

By using joules heating theory, $P = I^{2} R$ ,

$or,$ $P = 2^{2} * 75$

$or,$ $P= 4*75$

$or$ , $P= 300 W$ .

So the power dissipated by the resistor in the circuit is 300 W.

7 0

3 years ago

\nljbvjlsfhglskhg;skgh;sfjs;ojsr;ogjs;dofhs;dkhs;lifhns

vladimir1956 [14]

That is nonsense, also known as jibberish

8 0

3 years ago

Suppose you are an astronomer, and a child asks you to explain the light given off by stars. In your own words, describe the rel

Nadya [2.5K]

Well, i would tell the kid that light is electromagnetic radiation which is visible. The frequency and wavelength is just right so the radiation becomes visible, and that visible radiation goes away from the star and reaches earth It also depends on the age of the child and the comprehension level

3 0

3 years ago

Read 2 more answers

At the beach in San Francisco (0 meters) the pressure of the atmosphere is 101.325 kPa

Korvikt [17]

Answer:

$P = -\frac{17978}{1609344}(h)+101.325$

Explanation:

Given

$h = height$

$P = Pressure$

$(h_1,P_1) = (0,101.325)$

$(h_2,P_2) = (1609.344 ,83.437 )$

Required

Determine the linear equation for P in terms of h

First, we calculate the slope/rate (m);

The following formula is used:

$m = \frac{P_2 - P_1}{h_2 - h_1}$

Substitute values for P's and h's

$m = \frac{83.347 - 101.325}{1609.344- 0}$

$m = \frac{-17.978}{1609.344}$

$m = -\frac{17.978}{1609.344}$

Multiply by 1000/1000

$m = -\frac{17.978 * 1000}{1609.344*1000}$

$m = -\frac{17978}{1609344}$

The equation is then calculated using:

$P - P_1 = m(h - h_1)$

Substitute values for m, h1 and P1

$P - P_1 = m(h - h_1)$

$P - 101.325 = -\frac{17978}{1609344}(h - 0)$

$P - 101.325 = -\frac{17978}{1609344}(h)$

Make P the subject

$P = -\frac{17978}{1609344}(h)+101.325$

<em>The above is the required linear equation</em>

8 0

3 years ago

A centrifuge has an angular velocity of 3,000 rpm, what is the acceleration (in unit of the earth gravity) at a point with a rad

Anna71 [15]

Answer:

$a_{r} = 1006.382g \,\frac{m}{s^{2}}$

Explanation:

Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:

$a_{r} = \omega^{2}\cdot R$

Where:

$\omega$ - Angular speed, measured in radians per second.

$R$ - Radius of rotation, measured in meters.

The angular speed is first determined:

$\omega = \frac{\pi}{30}\cdot \dot n$

Where $\dot n$ is the angular speed, measured in revolutions per minute.

If $\dot n = 3000\,rpm$ , the angular speed measured in radians per second is:

$\omega = \frac{\pi}{30}\cdot (3000\,rpm)$

$\omega \approx 314.159\,\frac{rad}{s}$

Now, if $\omega = 314.159\,\frac{rad}{s}$ and $R = 0.1\,m$ , the resultant acceleration is then:

$a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)$

$a_{r} = 9869.588\,\frac{m}{s^{2}}$

If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:

$a_{r} = 1006.382g \,\frac{m}{s^{2}}$

6 0

3 years ago