The work function of a metallic surface is $5.01 \, eV$. The photo-electrons are emitted when light of wavelength $2000 \, \mathring{A}$ falls on it. The potential difference applied to stop the fastest photo-electrons is ............... $volt$ $[h = 4.14 \times 10^{-15} \, eV \cdot s]$

  • A
    $1.2$
  • B
    $2.24$
  • C
    $3.6$
  • D
    $4.8$

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Similar Questions

When the electromagnetic radiations of frequencies $4 \times 10^{15} \,Hz$ and $6 \times 10^{15} \,Hz$ fall on the same metal in different experiments,the ratio of maximum kinetic energy of electrons liberated is $1: 3$. The threshold frequency for the metal is ............... $\times 10^{15} Hz$.

The stopping potential for a source of wavelength $4000\, \mathring{A}$,when kept at a distance of $10\, \text{cm}$,is $1.5\, \text{V}$. If now the distance of the source is increased to $20\, \text{cm}$,then the stopping potential will be ............... $\text{V}$.

Photons of energy $6 eV$ are incident on a metal surface whose work function is $4 eV$. The minimum kinetic energy of the emitted photo-electrons will be ........... $eV$.

$A$ photon of energy $E$ ejects photoelectrons from a metal surface whose work function is $W_0$. If this electron enters into a uniform magnetic field with induction $B$ in a direction perpendicular to the field and describes a circular path of radius $r$,then the radius is given by

The figure shows the plot of stopping potential $(V_0)$ versus $(1/\lambda)$ for three different metals. If $\phi$ is the work function,then which of the following is correct?

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