Light of incident frequency $2$ times the threshold frequency is incident on a photosensitive material. If the incident frequency is made $\left(\frac{1}{3}\right)^{\text{rd}}$ and intensity is doubled,then the photoelectric current will

  • A
    decrease.
  • B
    increase.
  • C
    be halved.
  • D
    be zero.

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Similar Questions

$A$ mercury lamp is a convenient source for studying the frequency dependence of photoelectric emission,as it provides a number of spectral lines ranging from the $UV$ to the red end of the visible spectrum. In our experiment with a rubidium photocell,the following lines from a mercury source were used:
$\lambda_1 = 3650 \,\mathring{A}, \lambda_2 = 4047 \,\mathring{A}, \lambda_3 = 4358 \,\mathring{A}, \lambda_4 = 5461 \,\mathring{A}, \lambda_5 = 6907 \,\mathring{A}$
The stopping voltages,respectively,were measured to be:
$V_{01} = 1.28 \,V, V_{02} = 0.95 \,V, V_{03} = 0.74 \,V, V_{04} = 0.16 \,V, V_{05} = 0 \,V$
Determine the value of Planck's constant $h$,the threshold frequency,and the work function for the material.

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When a certain metal surface is illuminated with light of frequency $v$,the stopping potential for the photoelectric current is $V_s$. When the same surface is illuminated by a light of frequency $v/2$,the stopping potential is $V_s/4$. The threshold frequency of photoelectric emission is

$A$ light of wavelength $\lambda$ and intensity $I$ falls on a photosensitive material. If $N$ photoelectrons are emitted,each with kinetic energy $E$,then:

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