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

$A$ photosensitive metallic surface has a work function $\phi$. If a photon of energy $3 \phi$ falls on the surface,the electron is emitted with a maximum velocity of $6 \times 10^6 \ m/s$. When the photon energy is increased to $9 \phi$,the maximum velocity of the photoelectrons will be:

In the photoelectric effect,the stopping potential $(V_0)$ versus frequency $(\nu)$ curve is plotted. ($h$ is Planck's constant and $\phi_0$ is the work function of the metal)
$(A)$ $V_0$ versus $\nu$ is linear.
$(B)$ The slope of the $V_0$ versus $\nu$ curve $= \frac{\phi_0}{h}$.
$(C)$ Planck's constant $h$ is related to the slope of the $V_0$ versus $\nu$ line.
$(D)$ The value of the electric charge of an electron is not required to determine $h$ using the $V_0$ versus $\nu$ curve.
$(E)$ The work function can be estimated without knowing the value of $h$.
Choose the correct answer from the options given below:

Light of two different frequencies whose photons have energies $1 \text{ eV}$ and $2.5 \text{ eV}$ respectively,successively illuminate a metallic surface whose work function is $0.5 \text{ eV}$. The ratio of the maximum speeds of the emitted electrons will be:

When a piece of metal is illuminated by a monochromatic light of wavelength $\lambda$,the stopping potential is $3 V_{s}$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential becomes $V_{s}$. The value of the threshold wavelength for photoelectric emission is:

$A$ certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for the photoelectric current for this light is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is:

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