Dual Nature of Radiation and Matter

An electron (of mass $\mathrm{m}$ ) and a photon have the same energy $\mathrm{E}$ in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is ( $c=$  speed of light in vacuum)

An electron (mass m) with an initial velocity $\overrightarrow{v}=v_0\widehat{i}\left(v_0>0\right)$  is moving in an electric field $\overrightarrow{E}=-E_0\widehat{i}\left(E_0>0\right)$  where E₀ is constant. If at t = 0 de Broglie wavelength is ${\lambda }_0=\frac{h}{m{v}_0}$ , then its de Broglie wavelength after time t is given by

The ratio of wavelengths of proton and deuteron accelerated by potential V_p and V_d is $1:\sqrt[]{2}$ . Then, the ratio of V_p to V_d will be

An a particle and a proton are accelerated form rest through the same potential difference. The ratio of linear momenta acquired by above two particals will be:

An a particle and a proton are accelerated form rest through the same potential difference. The ratio of linear momenta acquired by above two particals will be:

The equation $\lambda =\frac{1.227}{x}nm$  can be used to find the de-Brogli wavelength of an electron. In this equation x stands for:

Where  m = mass of electron     

P = momentum of electron

K = kinetic energy of electron

V = Accelerating potential in volts for electron

A beam of electromagnetic radiation of intensity $6.4*{10}^{-5}\mathrm{W}/{\mathrm{c}\mathrm{m}}^2$  is comprised of wavelength, $\lambda =310\mathrm{n}\mathrm{m}$ . It falls normally on a metal (work function $\phi =2\mathrm{e}\mathrm{V}$  ) of surface area $1{\mathrm{c}\mathrm{m}}^2$ . If one in ${10}^3$  photons ejects an electron, total number of electrons ejected in is ${10}^x.\left(h_c=1240\mathrm{e}\mathrm{V}\mathrm{n}\mathrm{m},1\mathrm{e}\mathrm{V}=1.6*{10}^{-19}\mathrm{J}\right)$ , then $\mathrm{x}$  is     

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 speed of emitted electrons for the two frequencies respectively will be

The electric field at a point associated with a light wave is given by

E = 200 $\left[sin\left(6*10^{15}\right)t+sin\left(9*10^{15}\right)t\right]V{m}^{-1}$

Given : h = 4.14 * 10^-15 eVs

If the light fall 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: