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Dual Nature of Radiation and Matter Ncert Solutions Physics Class 12th Physics Ncert Solutions Class 12th Physics Class 12th Dual Nature of Radiation and Matter

11.25 Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.

(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radio waves of wavelength 500 m.

(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive ( 10^–10 W m^-2). Take the area of the pupil to be about 0.4 cm², and the average frequency of white light to be about 6 × 10¹⁴ Hz.

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Payal Gupta | Contributor-Level 10

4 months ago

11.25 The power of the medium wave transmitter, P = 10 kW = 10 $\times 10^{3}$ W = $10^{4}$ J/s

Hence energy emitted by the transmitter per second, E = $10^{4}$ J

Wavelength of the radio wave, $λ$ = 500 m

Planck’s constant, h = 6.626 $\times 10^{- 34}$ Js

Speed of light, c = 3 $\times 10^{8}$ m/s

Energy of the wave is given as :

$E_{w}$ = $\frac{h c}{λ}$ = $\frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{500}$ = 3.98 $\times 10^{- 28}$ J

Let n be number of photons emitted by the transmitter. Hence, total energy transmitted is given by:

n $E_{w}$ = E

n = $\frac{E}{E_{w}}$ = $\frac{10^{4}}{3.98 \times 10^{- 28}}$ = 2.52 $\times 10^{31}$

Intensity of light perceived by the human eye, I = $10^{- 10}$ W $m^{- 2}$

Area of the pupil, A = 0.4 ${c m}^{2}$ = 0.4&

...more

11.25 The power of the medium wave transmitter, P = 10 kW = 10 $\times 10^{3}$ W = $10^{4}$ J/s

Hence energy emitted by the transmitter per second, E = $10^{4}$ J

Wavelength of the radio wave, $λ$ = 500 m

Planck’s constant, h = 6.626 $\times 10^{- 34}$ Js

Speed of light, c = 3 $\times 10^{8}$ m/s

Energy of the wave is given as :

$E_{w}$ = $\frac{h c}{λ}$ = $\frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{500}$ = 3.98 $\times 10^{- 28}$ J

Let n be number of photons emitted by the transmitter. Hence, total energy transmitted is given by:

n $E_{w}$ = E

n = $\frac{E}{E_{w}}$ = $\frac{10^{4}}{3.98 \times 10^{- 28}}$ = 2.52 $\times 10^{31}$

Intensity of light perceived by the human eye, I = $10^{- 10}$ W $m^{- 2}$

Area of the pupil, A = 0.4 ${c m}^{2}$ = 0.4 $\times 10^{- 4} m^{2}$

Frequency of white light, $ν$ = 6 $\times 10^{14}$ Hz

Planck’s constant, h = 6.626 $\times 10^{- 34}$ Js

Energy emitted by photon is given as:

$E = h ν$ = 6.626 $\times 10^{- 34} \times 6 \times 10^{14}$ = 3.9756 $\times 10^{- 19}$ J

If n is total number of photons falling per second, per unit area of the pupil,

Total energy per unit area of n photons is = n $\times$ 3.9756 $\times 10^{- 19}$ J = Intensity of light

n $\times$ 3.9756 $\times 10^{- 19}$ = $10^{- 10}$

n = 2.52 $\times 10^{8}$

Total number of photons entering the pupil per second is given as:

$n_{A}$ = n $\times A$ = 2.52 $\times 10^{8} \times$ 0.4 $\times 10^{- 4}$ = 10.06 $\times 10^{3}$ /s

The number is not as large as the one found in problem (a), but it is large enough for the human eye to never see the individual photons.

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Given below are two statements:

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Statement II : If electrons have wave nature, they can interfere and show diffraction.

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Based on theory

Radiation from hydrogen gas excited to first excited state is used for illuminating certain metallic plate. When the same plate is exposed to the radiation from some unknown hydrogen like gas excited to the same level it is found that de-Broglie wavelength of fastest photoelectron has decreased by 2.3 times. It is given that energy corresponding to longest wavelength of Lyman series of the unknown gas is 3 times the ionization energy of hydrogen gas (13.6eV). The work function (in eV) of the metallic plate is ______. [Take (2.3)² = 5.25]

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z² × (13.6) (1 - ¼) = 3 × (13.6)
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h/√2mk? = (1/2.3) × h/√2mk?
=> k? = (2.3)²k? = 5.25k? (ii)
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A metal exposed to light of wavelength 800nm and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500nm is used. The work function of the metal is: (Take hc = 1230 eV –nm).

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$_{} \frac{}{}$ $K . E_{1} = \frac{1230}{800} - ?$ - (i)

$_{}_{} \frac{}{}$ $K . E_{2} = 2 K . E_{1} = \frac{1230}{500} - ?$ - (ii)

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When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is $υ_{1} .$ When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes $υ_{2} .$ If $υ_{2} = x υ_{1},$ the value of x will be __________.

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hu = hu₀ + K.E

Cases u = 2u₀

h2u₀ = hu₀ + K.E₁

K.E₁ = hu₀

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Now, cases 2

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$\frac{1}{2} m v_{2}^{2} = 4 h υ_{0}$

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Kindly consider the following

$f (x) = {\begin{matrix} \frac{2 x^{2} - 3 x + 2}{x - 2}, & x \neq 2 \\ 5, & x = 2 \end{matrix}$

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This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} f (x) = \frac{2 x^{2} - 3 x - 2}{x - 2} \\ = \frac{2 x^{2} - 4 x + x - 2}{x - 2} = \frac{2 x (x - 2) + 1 (x - 2)}{x - 2} \\ = \frac{(2 x + 1) (x - 2)}{x - 2} = 2 x + 1 \\ \underset{x \to 2^{-}}{l i m} f (x) = 2 x + 1 = \underset{h \to 0}{l i m} 2 (2 - h) + 1 = 4 + 1 = 5 \\ \underset{x \to 2^{+}}{l i m} f (x) = 2 x + 1 = \underset{h \to 0}{l i m} 2 (2 + h) + 1 = 4 + 1 = 5 \\ \underset{x \to 2}{l i m} f (x) = 5 \\ A s \underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2}{l i m} f (x) = 5 \\ H e n c e, f (x) i s c o n t i n u o u s a t x = 2 . \end{array}$

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