Advertisements
Advertisements
Question
Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
Advertisements
Solution
Wave nature of radiation cannot explain the following:
(i) The instantaneous ejection of photoelectrons.
(ii) The existence of threshold frequency for a metal surface.
(iii) The fact that kinetic energy of the emitted electrons is independent of the intensity of light and depends upon its frequency.
Thus, the photoelectric effect cannot be explained on the basis of wave nature of light.
APPEARS IN
RELATED QUESTIONS
An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Sodium and copper have work function 2.3 eV and 4.5 eV respectively. Then, the ratio of the wavelengths is nearest to ______.
An electromagnetic wave of wavelength ‘λ’ is incident on a photosensitive surface of negligible work function. If ‘m’ mass is of photoelectron emitted from the surface has de-Broglie wavelength λd, then ______
A proton, a neutron, an electron and an α-particle have same energy. Then their de Broglie wavelengths compare as ______.
Relativistic corrections become necessary when the expression for the kinetic energy `1/2 mv^2`, becomes comparable with mc2, where m is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron?
- λ = 10 nm
- λ = 10–1 nm
- λ = 10–4 nm
- λ = 10–6 nm
Two particles A1 sand A2 of masses m1, m2 (m1 > m2) have the same de Broglie wavelength. Then ______.
- their momenta are the same.
- their energies are the same.
- energy of A1 is less than the energy of A2.
- energy of A1 is more than the energy of A2.
A proton and an α-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths λp and λa related to each other?
A particle of mass 4M at rest disintegrates into two particles of mass M and 3M respectively having non zero velocities. The ratio of de-Broglie wavelength of particle of mass M to that of mass 3M will be:
In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to ______.
