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6 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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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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6 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

...more

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6 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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6 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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6 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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6 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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6 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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7 months ago

Dual Nature of Radiation and Matter NCERT

A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to (a) H (b) H^1/2 (c) H⁰ (d) H^-1/2

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Pallavi Pathak

Contributor-Level 10

Explanation- velocity of a freely falling body is v= $\sqrt[]{2 g h}$

And $? = \frac{h}{m v} = \frac{h}{m \sqrt[]{2 g h}}$

$? = h$ ^-1

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7 months ago

Dual Nature of Radiation and Matter NEET

Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance 2 m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å.
(i) Estimate number of photons emitted by the bulb per second. [Assume no other losses]
(ii) Will there be photoelectric emission?

(iii) How much time would be required by the atomic disk to receive energy equal to work function (2 eV)?
(iv) How many photons would atomic disk receive within time duration calculated in (iii) above?
(v) Can you explain how photoelectric effect was

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Pallavi Pathak

Contributor-Level 10

Explanation-number of photon emitted per second n= $\frac{p}{\frac{h c}{?}} = \frac{p ?}{h c} = \frac{20 * 5000 * 10^{- 10}}{6.62 * 10^{- 34} * 3 * 10^{8}}$

$= 5 * 10^{19} s^{- 1}$

(ii) E=hc $/ ?$ = $\frac{6.62 * 10^{- 34} * 3 * 10^{8}}{5000 * 10^{- 10} * 1.6 * 10^{- 19}} = 2.48 e V$ this enegy is greater than 2 so emission is possible

(iii) work function $?$ = $\frac{p}{4 ? d^{2}} * ? r^{2} ? t$ = $?_{o}$

$? t$ = $\frac{4 ? d^{2}}{p r^{2}}$ = $\frac{4 * 2 * 16 * 1.6 * 10^{- 19} * 2^{2}}{20 * {(1.5 * 10^{- 10})}^{- 2}} = 28.4 s$

(iv) N= $\frac{n ? r^{2}}{4 ? d^{2}} * ? t$

= $\frac{5 * 10^{19} * {(1.5 * 10^{- 10})}^{2} * 28.4}{4 * ({2)}^{2}}$ =2

(v) as the time of emission is 11.04s so photoelectric is not spontaneous.

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