Given below are two statements : one is labelled as Assertion A and the 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, choose the most appropriate answer from the options given below.

The electric field at a point associated with a light wave is given by E = 200 [sin (6 $$\times$$ 1015)t + sin (9 $$\times$$ 1015)t] Vm$$-$$1 Given : h = 4.14 $$\times$$ 10$$-$$15 eVs If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be

Let K1 and K2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $$\lambda$$1 and $$\lambda$$2, respectively are incident on a metallic surface. If $$\lambda$$1 = 3$$\lambda$$2 then :

The de Broglie wavelengths for an electron and a photon are $$\lambda$$e and $$\lambda$$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 ?

An $$\alpha$$ particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths $$({\lambda }_{\alpha }:{\lambda }_{C12})$$ is :

An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are Ee and pe and that of photon are Eph and pph respectively. Which of the following is correct?

A metal surface is illuminated by a radiation of wavelength 4500 $$\stackrel{o}{A}$$. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90$${}^{\circ }$$ with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately :

A proton, a neutron, an electron and an $$\alpha$$ particle have same energy. If $$\lambda$$p, $$\lambda$$n, $$\lambda$$e and $$\lambda$$a are the de Broglie's wavelengths of proton, neutron, electron and $$\alpha$$ particle respectively, then choose the correct relation from the following :

Given below are two statements : Statement I : Davisson-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 :

The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :

A source of monochromatic light liberates 9 $$\times$$ 1020 photon per second with wavelength 600 nm when operated at 400 W. The number of photons emitted per second with wavelength of 800 nm by the source of monochromatic light operating at same power will be :

A metal exposed to light of wavelength $$800\mathrm{nm}$$ and and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength $$500\mathrm{nm}$$ is used. The workfunction of the metal is : (Take hc $$=1230\mathrm{eV}-\mathrm{nm}$$ ).

The ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1 : $$\sqrt{2}$$. Then the ratio of Vp to Vd will be :

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

An electron (mass $$\mathrm{m}$$) with an initial velocity $$\vec{v}=v_0\hat{i}\left(v_0>0\right)$$ is moving in an electric field $$\vec{E}=-E_0\hat{i}\left(E_0>0\right)$$ where $$E_0$$ is constant. If at $$\mathrm{t}=0$$ de Broglie wavelength is $${\lambda }_0=\frac{h}{m{v}_0}$$, then its de Broglie wavelength after time t is given by

With reference to the observations in photo-electric effect, identify the correct statements from below : (A) The square of maximum velocity of photoelectrons varies linearly with frequency of incident light. (B) The value of saturation current increases on moving the source of light away from the metal surface. (C) The maximum kinetic energy of photo-electrons decreases on decreasing the power of LED (light emitting diode) source of light. (D) The immediate emission of photo-electrons out of metal surface can not be explained by particle nature of light/electromagnetic waves. (E) Existence of threshold wavelength can not be explained by wave nature of light/ electromagnetic waves. Choose the correct answer from the options given below :

The equation $$\lambda =\frac{1.227}{x}\mathrm{nm}$$ can be used to find the de-Brogli wavelength of an electron. In this equation $$x$$ stands for : Where $$\mathrm{m}=$$ mass of electron $$\mathrm{P}=$$ momentum of electron $$\mathrm{K}=$$ Kinetic energy of electron $$\mathrm{V}=$$ Accelerating potential in volts for electron

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

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

An $$\alpha$$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be: