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Alex787 [66]
3 years ago
12

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

Physics
2 answers:
Andrew [12]3 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]3 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 cannon sends a projectile towards a target a distance 1420 m away. The initial velocity makes an angle 35◦ with the horizontal
ss7ja [257]

Answer:

v_{o}=141.51m/s

Explanation:

From the exercise we know the final x distance, the angle which the projectile is being released and acceleration of gravity

x=1420m\\g=-9.8m/s^{2}

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x=v_{ox}t=v_{o}cos(35)t

Solving for v_{o}

v_{o}=\frac{x}{tcos(35)} =\frac{1420m}{tcos(35)} (1)

Now, if we analyze the equation of y-position we got

y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}

At the end of the motion y=0

0=v_{o}sin(35)t+\frac{1}{2}gt^{2}

Knowing the equation for v_{o} in (1)

0=\frac{1420}{tcos(35)}tsin(35)-\frac{1}{2}(9.8)t^{2}

\frac{1}{2}(9.8)t^{2}=1420tan(35)

Solving for t

t=\sqrt{\frac{2(1420tan(35))}{9.8} } =14.25s

Now, we can solve (1)

v_{o}=\frac{1420m}{(14.25s)cos(35)}=141.51m/s

6 0
3 years ago
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IrinaK [193]

Answer:

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Answer:0.302

Explanation:

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\mu =\frac{80}{27\times 9.8}

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