For a photosensitive material,the work function is $W_0$ and the stopping potential is $V$. What is the wavelength of the incident radiation? ($h=$ Planck's constant,$c=$ velocity of light,$e=$ electronic charge)

  • A
    $\frac{h^2 c^2}{W_0+eV}$
  • B
    $\frac{hc}{W_0}$
  • C
    $\frac{hcV}{W_0}$
  • D
    $\frac{hc}{W_0+eV}$

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Similar Questions

In the case of the photoelectric effect,the graph of measured stopping potential $(V_0)$ against the frequency $(\nu)$ of incident light is a straight line. The slope of this line multiplied by the charge of an electron $(e)$ gives:

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:

When the energy of incident radiation is increased by $20 \%$,the kinetic energy of the photoelectrons emitted from a metal surface increases from $0.5 \ eV$ to $0.8 \ eV$. The work function of the metal is (in $eV$)

When light of wavelength $\lambda$ falls on the emitter of a photocell,the maximum speed of the emitted photoelectrons is $V$. If the incident wavelength is changed to $\frac{2\lambda}{3}$,the maximum speed of the emitted photoelectrons will be:

The stopping potential of the photoelectrons from a photocell is:

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