
Light of wavelength $\lambda $ falls on a metal having work function $\dfrac{{hc}}{{{\lambda _0}}}$. Photoelectric effect will take place only if:
A) $\lambda \geqslant {\lambda _0}$
B) $\lambda \geqslant 2{\lambda _0}$
C) $\lambda \leqslant {\lambda _0}$
D) $\lambda < \dfrac{{{\lambda _0}}}{2}$
Answer
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Hint: Recall that the photoelectric effect is a phenomenon in which the electrons are ejected from the surface of a metal when the light is made to fall on the surface of the metal. These ejected electrons are known as photoelectrons. The electrons absorb energy and get ejected.
Complete step by step solution:
When light is incident on the surface of metal, the electrons absorb some amount of light. This is because some amount of light is used as energy and the rest of the energy is used to increase kinetic energy of the electrons ejected.
According to Einstein equation of photoelectric effect, the maximum kinetic energy of the electrons can be written as:
$K.E. = h\nu - h{\nu _0}$---(i)
The above equation can also be written as
$h\nu = h{\nu _0} + K.E.$
Given that the wavelength of the light is $ = \lambda $
The work function is $\phi = \dfrac{{hc}}{{{\lambda _0}}}$
The energy of the metal is $E = \dfrac{{hc}}{\lambda }$
The photoelectric effect will take place only if the energy of this effect will be greater than or equal to the work function of the metal.
$\therefore $ As per the condition of photoelectric effect, it can be written that
$ \Rightarrow \dfrac{{hc}}{\lambda } \geqslant \dfrac{{hc}}{{{\lambda _0}}}$
Or it can be written that
$\dfrac{1}{\lambda } \geqslant \dfrac{1}{{{\lambda _0}}}$
Or $\lambda \leqslant {\lambda _0}$
$\therefore $ Photoelectric effect will take place only if $\lambda \leqslant {\lambda _0}$
Option C is the right answer.
Note: It is important to remember that the work function of a metal is defined as the minimum amount of energy that is required to remove an electron from the surface of the metal. The kinetic energy of the incident light varies with the intensity. This means the energy increases if the intensity of light is increased. But the light that has low intensity also ejects electrons. It does so after some time so that the electrons acquire enough energy to get emitted.
Complete step by step solution:
When light is incident on the surface of metal, the electrons absorb some amount of light. This is because some amount of light is used as energy and the rest of the energy is used to increase kinetic energy of the electrons ejected.
According to Einstein equation of photoelectric effect, the maximum kinetic energy of the electrons can be written as:
$K.E. = h\nu - h{\nu _0}$---(i)
The above equation can also be written as
$h\nu = h{\nu _0} + K.E.$
Given that the wavelength of the light is $ = \lambda $
The work function is $\phi = \dfrac{{hc}}{{{\lambda _0}}}$
The energy of the metal is $E = \dfrac{{hc}}{\lambda }$
The photoelectric effect will take place only if the energy of this effect will be greater than or equal to the work function of the metal.
$\therefore $ As per the condition of photoelectric effect, it can be written that
$ \Rightarrow \dfrac{{hc}}{\lambda } \geqslant \dfrac{{hc}}{{{\lambda _0}}}$
Or it can be written that
$\dfrac{1}{\lambda } \geqslant \dfrac{1}{{{\lambda _0}}}$
Or $\lambda \leqslant {\lambda _0}$
$\therefore $ Photoelectric effect will take place only if $\lambda \leqslant {\lambda _0}$
Option C is the right answer.
Note: It is important to remember that the work function of a metal is defined as the minimum amount of energy that is required to remove an electron from the surface of the metal. The kinetic energy of the incident light varies with the intensity. This means the energy increases if the intensity of light is increased. But the light that has low intensity also ejects electrons. It does so after some time so that the electrons acquire enough energy to get emitted.
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