Question

When a metal surface is illuminated by light of wavelengths \[400{\rm{ nm \,and\,250 nm}}\], the maximum velocities of the photoelectrons ejected are v and 2v respectively. The work function of the metal is (h=Planck’s constant, c=velocity of Light in air).
A. \[2{\rm{ hc}} \times {\rm{1}}{{\rm{0}}^6}J\]
B. \[{\rm{1}}{\rm{.5 hc}} \times {\rm{1}}{{\rm{0}}^6}J\]
C. \[{\rm{hc}} \times {\rm{1}}{{\rm{0}}^6}J\]
D. \[{\rm{0}}{\rm{.5 hc}} \times {\rm{1}}{{\rm{0}}^6}J\]

Answer 1

Hint: The total energy of a photon is equal to the sum of the energy utilized to eject an electron and the maximum kinetic energy of electrons. By using this work function can be obtained.

Formula used:
The energy of the photon is given by the equation is given as:
\[E = h\upsilon = \dfrac{{hc}}{\lambda }\].
Where, \[h\] is the Plank constant, \[\upsilon \] is the frequency of incident light, c is the speed of light and \[\lambda \] is the wavelength.
Kinetic energy of photoelectrons is given as:
\[KE = E - \phi \]
Where E is the energy and \[\phi \] is the work function.
Also Kinetic energy is given as \[KE = \dfrac{1}{2}m{v^2}\]
Where m is the mass and v is the velocity.

Complete step by step solution:
As we know \[KE = \dfrac{{hc}}{\lambda } - \phi \]
For the maximum velocities of the photoelectrons ejected are v,
\[\dfrac{1}{2}m{v^2} = \dfrac{{hc}}{{{\lambda _1}}} - \phi \] ……. (1)
For the maximum velocities of the photoelectrons ejected are 2v,
\[\dfrac{1}{2}m{(2v)^2} = \dfrac{{hc}}{{{\lambda _2}}} - \phi \]
\[\Rightarrow \dfrac{1}{2}m(4{v^2}) = \dfrac{{hc}}{{{\lambda _2}}} - \phi \]
 \[\Rightarrow {\rm{ 4}}\left( {\dfrac{1}{2}m{v^2}} \right) = \dfrac{{hc}}{{{\lambda _2}}} - \phi \]

Put equation (1) in above equation,
\[{\rm{4}}\left( {\dfrac{{hc}}{{{\lambda _1}}} - \phi } \right) = \dfrac{{hc}}{{{\lambda _2}}} - \phi \]
\[\Rightarrow \dfrac{{4hc}}{{{\lambda _1}}} - \dfrac{{hc}}{{{\lambda _2}}} = 4\phi - \phi \\
\Rightarrow {\rm{ }}\phi {\rm{ = }}\dfrac{1}{3}\left( {\dfrac{{4hc}}{{{\lambda _1}}} - \dfrac{{hc}}{{{\lambda _2}}}} \right)nm\\
\Rightarrow {\rm{ }}\phi {\rm{ = }}\dfrac{{hc}}{3}\left( {\dfrac{4}{{{\lambda _1}}} - \dfrac{1}{{{\lambda _2}}}} \right) \times {10^9}m\]

Substituting the values
\[\Rightarrow {\rm{ }}\phi {\rm{ = }}\dfrac{{hc}}{3}\left( {\dfrac{4}{{{\lambda _1}}} - \dfrac{1}{{{\lambda _2}}}} \right) \times {10^9}\\
\Rightarrow {\rm{ }}\phi {\rm{ = }}\dfrac{{hc}}{3}\left( {\dfrac{4}{{400}} - \dfrac{1}{{250}}} \right) \times {10^9}\\
\Rightarrow {\rm{ }}\phi {\rm{ = }}\dfrac{{hc}}{{500}} \times {10^9}\\
\therefore {\rm{ }}\phi {\rm{ = }} 2{\rm{ hc}} \times {\rm{1}}{{\rm{0}}^6}J\]
 Therefore, the work function of the metal is \[2{\rm{ hc}} \times {10^6}J\].

Hence option A is the correct answer.

Note: The photoelectric effect occurs when electrons are released from a metal surface when sufficient frequency light is impressed on it. Einstein proposed that light acted like a particle, with each particle containing energy known as a photon. The energy of a wave is proportional to its amplitude, according to wave physics.