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# The threshold wavelength for photoelectric emission from a material is $5200A$ . Photo-electrons will be emitted when this material is illuminated with monochromatic radiation from a(A) $50watt$ infrared lamp(B) $1watt$ infrared (C) $1watt$ ultraviolet lamp(D) $60watt$ infrared lamp

Last updated date: 20th Jun 2024
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Hint: When light falls on the surface of some metals like Sodium then their electrons gain some energy from the light and start emitting with some kinetic energy, this phenomenon is called photoelectric emission. To start this emission the maximum wavelength of light which is needed is called threshold wavelength.

The photoelectric emission will take place if the energy of the coming light is greater than a minimum required energy which is also called the threshold energy.
So, the kinetic energy of the electrons$\left( {K.E} \right)$ will be the difference of the energy of the coming light$\left( E \right)$ and the threshold energy $\left( {{E_0}} \right)$
We can write,
$K.E = E - {E_0}$ = $h\nu - h{\nu _0}$ $= \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}}$
Where $h$ is the Planck’s constant, $\nu$ and $\lambda$ are respectively the frequency and wavelength of the coming light, ${\nu _0}$ and ${\lambda _0}$ are respectively threshold frequency and threshold wavelength.
For the photoelectric emission $K.E$$>$ 0
Therefore
$\Rightarrow \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}} > 0$
$\Rightarrow \dfrac{{hc}}{\lambda } > \dfrac{{hc}}{{{\lambda _0}}}$
$\Rightarrow \lambda < {\lambda _0}$
According to the question, ${\lambda _0} = 5200A$
Hence $\lambda < 5200A$
As the range of the wavelength of U.V light is lesser than$4000A$ and the range of the wavelength of infrared light is greater than $7800A$ , the U.V light satisfies our condition of $\lambda < 5200A$. Hence the material is illuminated by the U.V light.

The correct option is C.

Note: We can also solve this problem with frequency. In that case the frequency of the coming light must be greater than the threshold frequency. On the other hand the photoelectric emission doesn’t depend on the power of the light. Only the number of electrons emitted during the emission depends on how intense the power of the light is.