Question

Photoelectric work function of metal is 3.2 eV. Find the threshold wavelength.

Answer 1

Hint: The photoelectric effect formula should be used to solve this problem. The photoelectric effect formula gives the relation between the Planck’s constant, the speed of light, the wavelength of that wave and the photoelectric work function. The unit of the work function should be converted to Joules.
Formula used:
\[\dfrac{hc}{\lambda }=\phi \]

Complete step by step answer:
From the given information, we have the data as follows.
Photoelectric work function of metal is 3.2 eV.
The photoelectric effect formula gives the relation between the Planck’s constant, the speed of light, the wavelength of that wave and the photoelectric work function. The mathematical representation of the same is given as follows.
\[\dfrac{hc}{\lambda }=\phi \]
Where h is the Planck's constant, c is the speed of light, \[\lambda \]is the wavelength and \[\phi \]is the work function.
The values of the constants are known, that is, the constants are: the Planck's constant and the speed of light.
The Planck's constant is, \[h=6.63\times {{10}^{-34}}\]
The speed of light is, \[c=3\times {{10}^{8}}{m}/{s}\;\]
Consider the formula
\[\dfrac{hc}{\lambda }=\phi \]
Substitute the values of the Planck's constant, the speed of light and the work function in the above equation.
\[\dfrac{(6.63\times {{10}^{-34}})(3\times {{10}^{8}})}{\lambda }=3.2\,eV\]
Now, multiply the charge on the electron with the work function to convert its unit from electron volt to Joules.
\[\dfrac{(6.63\times {{10}^{-34}})(3\times {{10}^{8}})}{\lambda }=3.2\,\times 1.6\times {{10}^{-19}}J\]
Continue further computation.
\[\lambda =\dfrac{(6.63\times {{10}^{-34}})(3\times {{10}^{8}})}{3.2\,\times 1.6\times {{10}^{-19}}}\]
Thus, the value of the threshold wavelength is,
\[\lambda =3.86\times {{10}^{-7}}m\]
\[\therefore \] The value of the threshold wavelength, for the photoelectric work function of metal being 3.2 eV is, \[3.86\times {{10}^{-7}}m\].

Note: The unit conversion from the electron volts to joules should be done, before starting the computation. The reason being, to obtain the wavelength in terms of meters. The value of Planck's constant and the speed of light should be known to solve this problem.