The electric field associated with a light wave is given by $E=E_0\sin [(1.57\times 10^{7}m{s}^{-1})(ct-x)]E\; =\; E\_0\backslash sin[(1.57\backslash times10^\{7\}ms^\{-1\})(ct-x)]$. Find the stopping potential when this light is used in an experiment on photoelectric effect with the similar having work function 1.9 eV

The work function of potassium is 2.24 eV. The region of the electro spectrum which liberates the electron from potassium is

An electron accelerated by a p.d of V possess a de Broglie wavelength $\lambda \backslash lambda$. If the accelerating potential is increased by a factor 4. The de Broglie wavelength of the electron will

Relative to the original photon the photon that emerges after a collision with an electron has

An electron, proton, and an alpha particle all have the same kinetic energy. The de Broglie wavelength is

A particle is moving three times as fast as an electron. The ratio of the de- Broglie wavelength of the particle to that of the electron is $1.813\times 10^{-4}1.813\backslash times10^\{-4\}$. Calculate the particle's mass and identify the particle. Mass of electron = $9.11\times 10^{-31}9.11\backslash times10^\{-31\}$ kg.

The energy of a proton is equal to the kinetic energy of a photon. The energy of the photon is E. Let ${\lambda }_1\backslash lambda\_1$ be de Broglie wavelength of proton and ${\lambda }_2\backslash lambda\_2$ the wavelength of photon. Then we have

The de Broglie wavelength of electron in ground state of hydrogen atom is (given that the radius of the first orbit of hydrogen atom’s 0.53 Å)

The work function for aluminium surface is 4.2 eV. The cut off wavelength for photoelectric effect is

A metal surface ejects electrons when green light is incident on it but no emission take place when yellow light is incident on it. The electrons will be emitted when the surface has