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Dual Nature of Radiation and Matter

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2 months ago

Dual Nature of Radiation and Matter Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

An ideal fluid of density 800kgm^-3, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to $\frac{a}{2} .$ The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is $\frac{\sqrt[]{x}}{6} m s^{- 1}$ for x =________. (Given g = 10 ms^-²)

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Payal Gupta

Contributor-Level 10

$ρ = 800 k g / m^{3}$

P₁ – P₂ = 4100 Pa

Bernoulli's question b/w (1) & (2)

$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + z_{1} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + z_{2}$

$\frac{P_{1} - P_{2}}{ρ g} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{v_{2}^{2}}{2 g} + 0$ -(1)

Equation of continuity

A₁v₁ = A₂v₂

v₂ = 2v₁-(2)

from (1) & (2)

$\Rightarrow \frac{P_{1} - P_{2}}{ρ g} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{{(2 v_{1})}^{2}}{2 g}$

$\frac{4100}{800 * 10} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{4 v_{1}^{2}}{2 g}$

$\frac{12100}{8000} = \frac{3 v_{1}^{2}}{2 g}$

$\Rightarrow v_{1}^{2} = \frac{2 * 10}{3} * \frac{12100}{8000}$

$= \frac{2}{3} * \frac{121}{8}$

x = 363

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2 months ago

Dual Nature of Radiation and Matter physics ncert exemplar solutions class 12th chapter two

A heat engine operates with the could reservoir at temperature 324K. The minimum temperature of the hot reservoir, if the heat engine takes 300J heat from the hot reservoir and delivers 180J heat to the cold reservoir per cycle, is__________K.

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2 months ago

Dual Nature of Radiation and Matter physics ncert exemplar solutions class 12th chapter two

A metal surface is illuminated by a radiation of wavelength 4500 . The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90° with the magnetic field. If it starts revolving in a circular path of radius 2nm, the work function of the metal is approximately:

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Vishal Baghel

Contributor-Level 10

$k = \frac{h c}{λ} - ?$

As radius, r = $\frac{m v}{e B} = \frac{\sqrt[]{2 k m}}{e B}$

$\begin{array}{l} \Rightarrow {(r e B)}^{2} = 2 k m \\ \Rightarrow ? = \frac{h c}{λ} - \frac{{(r e B)}^{2}}{2 m} \end{array}$

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2 months ago

Dual Nature of Radiation and Matter Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

An electron with speed $υ$ and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are E_e and p_e and that of photon are E_ph and p_ph respectively. Which of the following is correct?

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Payal Gupta

Contributor-Level 10

$λ_{e} = \frac{h}{m v} = \frac{h}{p_{e}}$ - (1)

$λ_{P h} = \frac{h}{P_{P h}}$ - (2)

A/C to question

$λ_{e} = λ_{P h}$

$P_{e} = P_{P h} \Rightarrow \frac{P_{P}}{P_{P h}} = 1$ - (3)

$k . E_{e} = E_{e} = \frac{1}{2} m v^{2}$

$= \frac{1}{2} P e v$ - (4)

$k . E_{P h} = E_{P h} = m c^{2}$

= mc c

q = P_Ph c- (5)

$(4) \div (5)$

$\frac{E_{e}}{E_{P h}} = \frac{v}{2 c}$

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2 months ago

Dual Nature of Radiation and Matter Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Given below are two statements : one is labelled as Assertion A and other is labelled as Reason R.

Assertion A : The photoelectric effect does not takes place, if the energy of the incident radiation is less than the work function of a metal.

Reason R : Kinetic energy of the photoelectrons is zero, if the energy of the incident radiation is equal to the work function of a metal.

In the light of the above statements, chose the most appropriate answer from t

...more

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alok kumar singh

Contributor-Level 10

To free the electron from metal surface minimum energy required, is equal to the work function of that metal.

So Assertion A, is correct

have = w₀ + K.E_max

If have = w₀

Þ K.E_max = 0

Hence reas

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3 months ago

Dual Nature of Radiation and Matter Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is x : y. The value of x is

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Payal Gupta

Contributor-Level 10

Case I :- $2 ? - ? = \frac{1}{2} m v_{1}^{2} . . . . . . . . (i)$

Case II :- $10 ? - ? = \frac{1}{2} m v_{2}^{2} . . . . . . . . (i i)$

$\Rightarrow \frac{1}{9} = \frac{v_{1}^{2}}{v_{2}^{2}}$

$\Rightarrow v_{1} : v_{2} = 1 : 3$

x = 1

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3 months ago

Dual Nature of Radiation and Matter Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

A parallel beam of light of wavelength 900nm and intensity 100Wm^-2 is incident on a surface perpendicular to the beam. The number of photons crossing 1cm² area perpendicular to the beam in one second is:

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Vishal Baghel

Contributor-Level 10

$E = I A$

$= 100 * 1 * 1 0^{- 4}$

$= 1 0^{- 2} W$

$E = n h \frac{c}{λ}$

$1 0^{- 2} = \overset{?}{n} * \frac{6.64 * 1 0^{- 34} * 3 * 1 0^{8}}{900 * 1 0^{- 9}}$

$\overset{?}{n} = \frac{1 0^{- 2} * 9 * 1 0^{- 7}}{6.64 * 1 0^{- 34} * 3 * 1 0^{8}}$

$\overset{?}{n} = 4.5 * 1 0^{16}$

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3 months ago

Dual Nature of Radiation and Matter Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted, in the two cases respectively, will be

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Vishal Baghel

Contributor-Level 10

$E_{1} = 5 ? - ? = 4 ? = \frac{1}{2} m v_{1}^{2}$

$E_{2} = 10 ? - ? = \frac{1}{2} m V_{2}^{2}$

$\Rightarrow \frac{V_{1}}{V_{2}} = \frac{\sqrt[]{\frac{8 ?}{m}}}{\sqrt[]{\frac{18 ?}{m}}} = \sqrt[]{\frac{8}{18}} = \frac{\sqrt[]{4}}{9} = \frac{2}{3}$

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3 months ago

Dual Nature of Radiation and Matter Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

THE de Broglie wavelength for an electron and a photon are $λ_{e} a n d λ_{p}$ respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two

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Payal Gupta

Contributor-Level 10

$λ_{e} = \frac{h}{P_{e}} = \frac{h}{\sqrt[]{2 m_{e} K_{e}}} \Rightarrow λ_{e}^{2} = \frac{h^{2}}{2 m_{e} k_{e}} \Rightarrow K_{e} = \frac{h^{2}}{2 m_{e} λ_{e}^{2}}$ - (i)

$λ_{P} = \frac{h}{P_{P}} = \frac{h c}{K_{P}} \Rightarrow K_{P} = \frac{h c}{λ_{P}}$ - (ii)

K_e = K_P

$\Rightarrow λ_{P} α λ_{e}^{2}$

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New answer posted

3 months ago

Dual Nature of Radiation and Matter Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

The kinetic energy of emitted electron is E when the right incident on the metal has wavelength $λ$ . To double the kinetic energy, the incident light must have wavelength:

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Vishal Baghel

Contributor-Level 10

$\frac{h c}{λ} - ? = E$ -(i)

$\frac{h c}{λ'} - ? = 2 E$ -(ii)

$h c (\frac{1}{λ'} - \frac{1}{λ}) = E$

$λ' = \frac{h c λ}{E λ + h c}$

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