Question

If the work function of a photo metal is 6.825 eV. Its threshold wavelength will be (\[c = 3 \times {10^8}m/s\])
A. \[1200\mathop A\limits^ \circ \]
B. \[1800\mathop A\limits^ \circ \]
C. \[2400\mathop A\limits^ \circ \]
D. \[3600\mathop A\limits^ \circ \]

Answer 1

Hint: The threshold wavelength of a photo metal is the wavelength to which the corresponding energy of the photon is sufficient to eject the electron from the surface of that photo metal.

Formula used:
\[K = \dfrac{{hc}}{\lambda } - \phi \]
Here K is the kinetic energy of the emitted electron, h is the planck's constant, c is the speed of light, \[\lambda \]is the wavelength of the photon and \[\phi \]is the work function of the metal.

Complete step by step solution:
The threshold wavelength of the metal is the wavelength corresponding to the minimum energy which is sufficient to overcome the attractive force which causes the valence electron to be bound to the shell of the atom of the metal. As the energy of the photon is inversely proportional to the wavelength, so for a minimum value of the energy, the wavelength should be the maximum allowed wavelength.

When the energy of the photon is equal to the work energy of the photo metal then the wavelength of the photon is called the threshold wavelength. For the work function, the kinetic energy of the ejected electron is zero (for the minimum condition for emission of an electron).

The work function is given as 6.825 eV.
The work function in unit of joule will be,
\[\phi = 6.825 \times 1.6 \times {10^{ - 19}}J\]
\[\Rightarrow \phi = 1.092 \times {10^{ - 18}}J\]
Using the energy formula, we get
\[\dfrac{{hc}}{{{\lambda _0}}} = \phi \], here \[{\lambda _0}\]is the threshold wavelength of the metal.
So,
\[\Rightarrow {\lambda _0} = \dfrac{{hc}}{\phi }\]
\[\Rightarrow {\lambda _0} = \dfrac{{\left( {6.626 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{1.092 \times {{10}^{ - 18}}}}m\]
\[\Rightarrow {\lambda _0} = 1.821 \times {10^{ - 7}}m\]
\[\Rightarrow {\lambda _0} = 1821 \times {10^{ - 10}}m\]
\[\therefore {\lambda _0} \approx 1800\mathop A\limits^0 \]
Hence, the threshold wavelength of the given photo metal is \[1800\mathop A\limits^ \circ \].

Therefore, the correct option is B.

Note: The frequency of the incoming radiation affects the kinetic energy of the released electron. The maximal kinetic energy of the emitting electrons occurs at frequencies larger than. The amount of photoelectrons released is a function of the radiation's incidence intensity.