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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

11.4 Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.

(a) Find the energy and momentum of each photon in the light beam,

(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and

(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

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Payal Gupta | Contributor-Level 10

5 months ago

11.4 Wavelength of the monochromatic light, $λ = 632.8 n m$ = 632.8 $\times 10^{- 9}$ m

Power emitted by laser, P = 9.42 mW = 9.42 $\times 10^{- 3}$ W

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

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

Mass of hydrogen atom, m = 1.66 $\times 10^{- 27}$ kg

The energy of each photon is given as, $E$ = $\frac{h c}{λ}$ = $\frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{632.8 \times 10^{- 9}}$ J = 3.141 $\times 10^{- 19}$ J

The momentum of each photon is given by $p$ = $\frac{h}{λ}$ = $\frac{6.626 \times 10^{- 34}}{632.8 \times 10^{- 9}}$ = 1.047 $\times 10^{- 27}$ kg-m/s

Number of photons arriving per second at a target irradiated by the beam = n

Assume that the beam has a uniform cross-section that is less than the target area.

Hence, the equation for p

...more

11.4 Wavelength of the monochromatic light, $λ = 632.8 n m$ = 632.8 $\times 10^{- 9}$ m

Power emitted by laser, P = 9.42 mW = 9.42 $\times 10^{- 3}$ W

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

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

Mass of hydrogen atom, m = 1.66 $\times 10^{- 27}$ kg

The energy of each photon is given as, $E$ = $\frac{h c}{λ}$ = $\frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{632.8 \times 10^{- 9}}$ J = 3.141 $\times 10^{- 19}$ J

The momentum of each photon is given by $p$ = $\frac{h}{λ}$ = $\frac{6.626 \times 10^{- 34}}{632.8 \times 10^{- 9}}$ = 1.047 $\times 10^{- 27}$ kg-m/s

Number of photons arriving per second at a target irradiated by the beam = n

Assume that the beam has a uniform cross-section that is less than the target area.

Hence, the equation for power can be written as, $P = n E$

n = $\frac{P}{E}$ = $\frac{9.42 \times 10^{- 3}}{3.141 \times 10^{- 19}}$ = 3.0 $\times 10^{16}$

Momentum of the hydrogen atom = momentum of the proton = 1.047 $\times 10^{- 27}$ kg-m/s

The momentum of the hydrogen atom , $p = m v$ , where m = mass of hydrogen atom and v = velocity

Hence, v = $\frac{p}{m}$ = $\frac{1.047 \times 10^{- 27}}{1.66 \times 10^{- 27}}$ = 0.631 m/s

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Given below are two statements:

Statement I : Davission-Germer experiment establishes, the wave nature of electrons.

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

In the light of the above statements choose the correct answer from the option given below:

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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)
Now, k? = E? - Φ
k? = E? - Φ = z²E? - Φ
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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)

from (i) & (ii)

$? = 0.615$ ev

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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$v_{1}^{2} = \frac{2 h υ_{0}}{m}$

$v_{1} = \sqrt[]{\frac{2 h υ_{0}}{m}}$ - (1)

Now, cases 2

h 5u₀ = hu₀ + k.E₂

k.E₂ = 4hu₀

$\frac{1}{2} m v_{2}^{2} = 4 h υ_{0}$

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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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