When light of frequency $8 \times 10^{14} \ Hz$ is incident on a surface,photoelectrons are emitted with a maximum speed of $7 \times 10^5 \ m/s$. The threshold frequency for this surface is ..........

  • A
    $2.32 \times 10^{14} \ Hz$
  • B
    $4.64 \times 10^{14} \ Hz$
  • C
    $4.64 \times 10^{16} \ Hz$
  • D
    $4.64 \times 10^{18} \ Hz$

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Similar Questions

When light of wavelength $\lambda$ is incident on a photosensitive surface,the stopping potential is $V$. When light of wavelength $3 \lambda$ is incident on the same surface,the stopping potential is $\frac{V}{6}$. Then the threshold wavelength for the surface is:

The maximum velocity of an electron emitted by light of wavelength $\lambda$ incident on the surface of a metal of work function $\phi$ is:
Where $h =$ Planck's constant, $m =$ mass of electron, and $c =$ speed of light.

Einstein's photoelectric equation states that ${E_k} = h\nu - \phi$. In this equation,${E_k}$ refers to:

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.

If a photocell is illuminated with a radiation of $1240 \mathring{A}$,the stopping potential is found to be $8 \text{ V}$. Then the work function of the emitter and the threshold wavelength are:

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