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DanielleElmas [232]
3 years ago
5

In a particular beam of radiation, which is traveling in a vacuum, the amounts of energy per second at an ultraviolet wavelength

of 300 nm and at an infrared wavelength of 800 nm are found to be equal. In this beam, how do the numbers of photons per second at each of these wavelengths compare?
(a) There will be equal numbers of photons at each of these wavelengths.
(b) There will be more UV photons than IR photons.
(c) It is not possible to tell because we do not know the frequencies of the light.
(d) There will be more IR photons than UV photons.
Physics
1 answer:
Darina [25.2K]3 years ago
8 0

Answer: d)

Explanation: In order to justify the answer we have to consider that the energy of photons directely depent on the frequency so the energy is inverselly dependent of the wavelegth.

If both beams have the same power, this means Energy/time so the number of photons per second must be different. As consequence a) is wrong as  b) since it is not posible since UV photon  have more energy that IR photons. c) It is no necessary know the frequency since the wavelength is related in the form:

c=λν  c is the speed of light, λ the wavelegth and ν the frequency.

d) Certainly will be more more IR photons than UV photons to get the same beam power.

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Answer:

The temperature after half an hour is 19.3002^{\circ}

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As per the question;

Metabolic rate of the person is 3.0105 J/h

Temperature, T = 19.30^{\circ}

Mass of the water, m_{w} = 1.2103 kg

Time duration, t = 0.5 h = 30 min = 180 s

Now,

Heat, Q = ms\Delta t

Thus heat transfer in half an hour:

Q = 3.0105\times 0.5 = 1.505 J

Now, the temperature of water after half an hour, T' is given by:

Q = m_{w}s\Delta T = ms(T' - T)

where

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1.505 = 1.21103\times 4186\times (T' - 19.30^{\circ})

T' = 19.3002 ^{\circ}

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The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A
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Answer:

2.1\times 10^{-12} c

Explanation:

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Thickness of membrane=1.1\times 10^{-8} m

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We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

C=\frac{k\epsilon_0 A}{d}

Substitute the values then we get

Capacitance between parallel plate capacitor=\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}

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