Light of frequency $$4{\nu _0}$$, is incident on the metal of the threshold frequency $${\nu _0}$$. The maximum kinetic energy of the emitted photoelectrons isA. $$3\,h{\nu _0}$$B. $$2\,h{\nu _0}$$C. $$\dfrac{3}{2}h{\nu _0} \\ $$D. $$\dfrac{1}{2}h{\nu _0}$$

Question

Light of frequency $$4{\nu _0}$$​, is incident on the metal of the threshold frequency $${\nu _0}$$​. The maximum kinetic energy of the emitted photoelectrons isA. $$3\,h{\nu _0}$$B. $$2\,h{\nu _0}$$C. $$\dfrac{3}{2}h{\nu _0} \\ $$D. $$\dfrac{1}{2}h{\nu _0}$$

Vedantu Content Team · Accepted Answer

Hint:The maximum kinetic energy of the ejected electron during photoelectric effect is the residual energy transferred by the incident photon to the electron after overcoming the work function of the metal surface. The  work function of the metal is constant. So, greater will be the photon energy, greater will be the kinetic energy of the emitted photoelectrons.Formula used:$$K = h\nu  - \phi $$Here K is the kinetic energy of the emitted electron, h is the planck's constant, c is the speed of light, $$\nu $$ is the frequency of the photon and $$\phi $$is the work function of the metal. Complete step by step solution:The wavelength that must be present in order to overcome the attractive attraction that holds the valence electron to the metal atom's shell is known as the threshold wavelength of the metal. The wavelength should be the highest permitted wavelength for the minimal value of the photon's energy since the photon's energy is inversely related to its wavelength.When the kinetic energy is zero, then the corresponding energy of the photon is called the work function of the metal. The frequency corresponding to the energy equivalent to the work function is called the threshold frequency and it is given for the metal is $${\nu _0}$$. The frequency of the incident light is $$4{\nu _0}$$.  Using the formula of maximum kinetic energy, we get$${K_{\max }} = 4h{\nu _0} - h{\nu _0}$$$$\therefore {K_{\max }} = 3\,h{\nu _0}$$Hence, the maximum kinetic energy of the ejected electron is $$3\,h{\nu _0}$$.Therefore, the correct option is A.Note: The maximum kinetic energy of the photoelectron is proportional to the frequency of the incident photon. If the frequency of the incident photon is equal to the threshold frequency of the metal, then the maximum kinetic energy is zero as all the energy is used to make electrons come out of the metal surface.

Light of frequency \[4{\nu _0}\]​, is incident on the metal of the threshold frequency \[{\nu _0}\]​. The maximum kinetic energy of the emitted photoelectrons isA. \[3\,h{\nu _0}\]B. \[2\,h{\nu _0}\]C. \[\dfrac{3}{2}h{\nu _0} \\ \]D. \[\dfrac{1}{2}h{\nu _0}\]

Light of frequency \[4{\nu _0}\], is incident on the metal of the threshold frequency \[{\nu _0}\]. The maximum kinetic energy of the emitted photoelectrons is
A. \[3\,h{\nu _0}\]
B. \[2\,h{\nu _0}\]
C. \[\dfrac{3}{2}h{\nu _0} \\ \]
D. \[\dfrac{1}{2}h{\nu _0}\]