Question

The work function for tungsten and sodium are 4.5 eV and 2.3 eV respectively. If the threshold wavelength \[\lambda \] for sodium \[{\rm{5460}}\mathop {\rm{A}}\limits^{\rm{0}} \] , find the value of \[\lambda \] for tungsten.
A. \[{\rm{5893}}\mathop {\rm{A}}\limits^{\rm{0}} \]
B. \[{\rm{10683}}\mathop {\rm{A}}\limits^{\rm{0}} \]
C. \[2791\mathop {\rm{A}}\limits^{\rm{0}} \]
D. \[528\mathop {\rm{A}}\limits^{\rm{0}} \]

Answer 1

Hint: When light strikes a metal surface, electrons known as photoelectrons are released. The photoelectric effect is the name given to this occurrence. The work function is defined as the amount of energy necessary to remove one electron from a metal surface. Work function may alternatively be defined as the features of the surface rather than the entire metal. As we know, for photoemission to occur, the energy of the incident light or photons must be greater than the work function of the metal.

Formula Used:
To find the work function we have,
\[W = \dfrac{{hc}}{\lambda }\]
Where, h is Planck’s constant, c is the speed of light and \[\lambda \] is the threshold wavelength.

Complete step by step solution:
As we know that the work function is given by,
\[W = \dfrac{{hc}}{\lambda }\]
Here, \[W \propto \dfrac{1}{\lambda }\]
Therefore, for the tungsten and sodium, the work function and the wavelength can be,
\[\dfrac{{{W_N}}}{{{W_t}}} = \dfrac{{{\lambda _t}}}{{{\lambda _N}}}\]
We need to find the wavelength for the tungsten, for rearrange the above equation, and we get,
\[{\lambda _t} = \dfrac{{{W_N}{\lambda _N}}}{{{W_t}}}\]
\[\Rightarrow {\lambda _t} = \dfrac{{{W_N}{\lambda _N}}}{{{W_t}}}\]
Here, \[{W_t} = 4.5eV\], \[{W_N} = 2.3eV\] and the threshold wavelength of sodium is \[{{\rm{\lambda }}_N}{\rm{ = 5460}}\mathop {\rm{A}}\limits^{\rm{0}} \].
Now, substitute the values in the above equation we get,
\[{\lambda _t} = \dfrac{{2.3 \times 5460 \times {{10}^{ - 10}}}}{{4.5}}\]
\[\Rightarrow {\lambda _t} = 2790.6 \times {10^{ - 10}}m\]
\[\therefore {{\rm{\lambda }}_{\rm{t}}}{\rm{ = 2791}}\mathop {\rm{A}}\limits^{\rm{0}} \]
Therefore, the wavelength for tungsten is \[2791\mathop {\rm{A}}\limits^{\rm{0}} \].

Hence, Option C is the correct answer.

Note:Remember that, if the energy is given in eV, then we need to convert the given value of energy from eV to joules by multiplying the charge of an electron. But if we are finding the wavelength then it is not necessary to convert it into joules as shown in this problem. Moreover, metals contain many free electrons that are loosely bound to the atom. The typical value of work function varies from 2ev to 6eV.