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Photoelectric effect is considered one the most difficult concept.
15 Questions around this concept.
If and be the threshold wavelength and wavelength of the incident light, the velocity of the photoelectron ejected from the metal surface is :
For a metal surface, the work function is 3.5 eV. If the incident light has a wavelength of 400 nm, what is the kinetic energy of the emitted electrons?
Given below are two statements : one is labelled as Assertion and the other is labelled as Reason :
Assertion : In the photoelectric effect electrons are ejected from the metal surface as soon as the beam of light
of frequency greater than threshold frequency strikes the surface.
Reason : When the photon of any energy strikes an electron in the atom transfer of energy from the photon to
the electron takes place.
In the light of the above statements, choose the most appropriate answer from the options given below:
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Whenever a metal surface is exposed to light radiation of suitable energy or frequency, it was observed that some electrons get ejected from the metal surface. This phenomenon is called as Photoelectric effect and the ejected electrons are called as photoelectrons.
There were certain observation in the photoelectric effect experiment.
(1) There was a requirement of a minimum energy for each metal for the photoelectric effect to occur. This minimum energy is known as work function (W0) and it can be closely associated with the ionisation energy of the metal.
Mathematically, the work function, threshold frequency and threshold wavelength can be associated as
$\mathrm{W}_0=\mathrm{h} \nu_0=\frac{\mathrm{hc}}{\lambda_0}$
Note: hc is approximately equal to 2 x 10-25 J-m or 12400 eV-nm. (eV is the energy in electron volts)
(2) The number of electrons ejected is proportional to the intensity (brightness) of light striking the metal but does not depend upon the frequency of light.
(3) There was almost no time lag between the striking of light and ejection of photo electrons
(4) The kinetic energy of the ejected electrons (photoelectrons) depend upon the frequency of the light used.
Einstein's photoelectric equation
From conservation of energy
$\begin{gathered}E_p=W_0+K E \\ \therefore h \nu=h \nu_0+\frac{1}{2} m v^2\end{gathered}$
where
m is the mass of the electron
v is the velocity associated with the ejected electron.
h is planck’s constant.
v is frequency of photon,
v0 is threshold frequency of metal.
(5) The Kinetic energy of ejected photoelectron is also sometimes associated to Stopping Potential. It is defined as the minimum opposing potential applied due to which kinetic energy of electron becomes zero.
$\frac{1}{2} m v^2=e V_s$
where,
Vs = Stopping potential
e = Charge on electron
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