Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A) :$ Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electric potential to the photoemissive substance.
Reason $(R) :$ $A$ negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation.
In the light of the above statements, choose the most appropriate answer from the options given below:

  • A
    $(A)$ is false but $(R)$ is true.
  • B
    $(A)$ is true but $(R)$ is false.
  • C
    Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A).$
  • D
    Both $(A)$ and $(R)$ are true but $(R)$ is not the correct explanation of $(A).$

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Similar Questions

Given below are two statements:
Statement-$I$: The figure shows the variation of stopping potential $(V_0)$ with frequency $(v)$ for two photosensitive materials $M_1$ and $M_2$. The slope gives the value of $\frac{h}{e}$,where $h$ is Planck's constant and $e$ is the charge of an electron.
Statement-$II$: $M_2$ will emit photoelectrons of greater kinetic energy for incident radiation having the same frequency.
In the light of the above statements,choose the most appropriate answer from the options given below.

$A$ photon of energy $3.4 \ eV$ is incident on a metal surface with a work function of $2 \ eV$. The maximum kinetic energy of the emitted photoelectrons is .......... $eV$.

Ultraviolet light of wavelength $300 \ nm$ and intensity $1.0 \ W/m^2$ is incident on the surface of a photosensitive material. If $1\%$ of the incident photons produce photoelectrons,the number of photoelectrons emitted from an area of $1.0 \ cm^2$ of the surface is:

The de Broglie wavelength of the most energetic photoelectrons emitted from a photosensitive metal of work function $\phi$,when light of frequency $\nu$ is incident on it,is $\lambda$. Then $\nu =$ (where $h$ is Planck's constant and $m$ is the mass of the electron).

In a photo-emissive cell,with exciting wavelength $\lambda$,the maximum kinetic energy of the electron is $K$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the kinetic energy of the fastest emitted electron will be:

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