By the photoelectric effect,Einstein proved:

  • A
    $E = h\nu$
  • B
    $K.E. = \frac{1}{2}mv^2$
  • C
    $E = mc^2$
  • D
    $E = \frac{-Rhc^2}{n^2}$

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Similar Questions

$A$ metal surface having work function '$W_{0}$' emits photoelectrons when photons of energy '$E$' are incident on it. The electron enters a uniform magnetic field '$B$' in a perpendicular direction and moves in a circular path of radius '$r$'. Then '$r$' is equal to (where '$m$' and '$e$' are the mass and charge of the electron,respectively).

Light of frequency $8 \times 10^{15} \ Hz$ is incident on a substance of photoelectric work function $6.125 \ eV$. The maximum kinetic energy of the emitted photoelectrons is ........... $eV$.

$A$ light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of $2 \ \text{eV}$. If the same surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$, then the maximum kinetic energy of ejected electrons will be (The work function of the metal is $1 \ \text{eV}$). (in $\text{eV}$)

In a photoelectric experiment,ultraviolet light of wavelength $280 \, nm$ is used with a lithium cathode having a work function $\phi = 2.5 \, eV$. If the wavelength of incident light is switched to $400 \, nm$,find out the change in the stopping potential (in $V$).
$(h = 6.63 \times 10^{-34} \, J \cdot s, c = 3 \times 10^8 \, m/s)$

The frequency of incident light falling on a photosensitive material is doubled. The kinetic energy $(K.E.)$ of the emitted photoelectrons will be:

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