From the figure describing the photoelectric effect,we may infer correctly that:

  • A
    $Na$ and $Al$ both have the same threshold frequency.
  • B
    Maximum kinetic energy for both the metals depends linearly on the frequency.
  • C
    The stopping potentials are different for $Na$ and $Al$ for the same change in frequency.
  • D
    $Al$ is a better photosensitive material than $Na$.

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The maximum velocity of the photoelectrons emitted from the surface is $v$ when light of frequency $n$ falls on a metal surface. If the incident frequency is increased to $3n$,the maximum velocity of the ejected photoelectrons will be:

Radiation of wavelength $332 \ nm$ is incident on a metal surface having a work function of $1.07 \ eV$. The stopping potential required to stop the emission of photoelectrons from the metal surface is ............ $V$. $(h = 6.6 \times 10^{-34} \ J s, c = 3 \times 10^8 \ m/s, 1 \ eV = 1.6 \times 10^{-19} \ J)$

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The kinetic energy with which electrons are emitted from a metal surface due to the photoelectric effect is:

The slope of the graph between stopping potential and the frequency of incident radiation for the photoelectric effect is . . . . . . .

Photoelectrons are emitted when photons of energy $4.2 \text{ eV}$ are incident on a photosensitive metallic sphere of radius $10 \text{ cm}$ and work function $2.4 \text{ eV}$. The number of photoelectrons emitted before the emission is stopped is
$\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text{ SI unit; } e=1.6 \times 10^{-19} \text{ C}\right]$

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