Answer
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Hint: We need to find the work function. This can be done from Einstein’s equation which is given by $K{E_{\max }} = \dfrac{{hc}}{\lambda } - \phi $. This is a concept of modern physics.
Formula used: By Planck- Einstein relation:
$K{E_{\max }} = \dfrac{{hc}}{\lambda } - \phi $
Here,
$K{E_{\max }}$ is the maximum kinetic energy
$h$ is the Planck’s constant
$c$is the speed of light
$\phi $ is the work function
$\lambda $ is the wavelength
Complete step by step answer:
We already know from the above formula,
We just need to put the values:
For the situation when wavelength is $\lambda $,
$K{E_{\max }} = \dfrac{{hc}}{\lambda } - \phi $
And for the second case when wavelength is half the initial,
$K{E_{\max 2}} = \dfrac{{2hc}}{\lambda } - \phi $
But as given in the question:
$K{E_{\max 2}} = 3K{E_{\max 1}}$
So, \[\dfrac{{2hc}}{\lambda } - \phi = 3\left( {\dfrac{{hc}}{\lambda } - \phi } \right)\]
On solving further,
\[3\phi - \phi = \dfrac{{3hc}}{\lambda } - \dfrac{{2hc}}{\lambda }\]
\[\phi = \dfrac{{hc}}{{2\lambda }}\]
Hence, we obtain the work function to be $\dfrac{{hc}}{{2\lambda }}$.
We need to select the correct option.
The correct option is B.
Additional Note: Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. The higher the photon's frequency the higher is its energy.
Note: It should be kept in mind that Einstein's photo-electric equation is important. Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. This equation is known as the Planck-Einstein relation.
Since light is bundled up into photons, Einstein theorized that when a photon falls on the surface of a metal, the entire photon’s energy is transferred to the electron.
A part of this energy is used to remove the electron from the metal atom’s grasp and the rest is given to the ejected electron as kinetic energy. Electrons emitted from underneath the metal surface lose some kinetic energy during the collision. But the surface electrons carry all the kinetic energy imparted by the photon and have the maximum kinetic energy.
Formula used: By Planck- Einstein relation:
$K{E_{\max }} = \dfrac{{hc}}{\lambda } - \phi $
Here,
$K{E_{\max }}$ is the maximum kinetic energy
$h$ is the Planck’s constant
$c$is the speed of light
$\phi $ is the work function
$\lambda $ is the wavelength
Complete step by step answer:
We already know from the above formula,
We just need to put the values:
For the situation when wavelength is $\lambda $,
$K{E_{\max }} = \dfrac{{hc}}{\lambda } - \phi $
And for the second case when wavelength is half the initial,
$K{E_{\max 2}} = \dfrac{{2hc}}{\lambda } - \phi $
But as given in the question:
$K{E_{\max 2}} = 3K{E_{\max 1}}$
So, \[\dfrac{{2hc}}{\lambda } - \phi = 3\left( {\dfrac{{hc}}{\lambda } - \phi } \right)\]
On solving further,
\[3\phi - \phi = \dfrac{{3hc}}{\lambda } - \dfrac{{2hc}}{\lambda }\]
\[\phi = \dfrac{{hc}}{{2\lambda }}\]
Hence, we obtain the work function to be $\dfrac{{hc}}{{2\lambda }}$.
We need to select the correct option.
The correct option is B.
Additional Note: Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. The higher the photon's frequency the higher is its energy.
Note: It should be kept in mind that Einstein's photo-electric equation is important. Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. This equation is known as the Planck-Einstein relation.
Since light is bundled up into photons, Einstein theorized that when a photon falls on the surface of a metal, the entire photon’s energy is transferred to the electron.
A part of this energy is used to remove the electron from the metal atom’s grasp and the rest is given to the ejected electron as kinetic energy. Electrons emitted from underneath the metal surface lose some kinetic energy during the collision. But the surface electrons carry all the kinetic energy imparted by the photon and have the maximum kinetic energy.
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