Question

When the light of wavelength \[300{\text{ }}nm\] falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, however, the light of \[600nm\] wavelength is sufficient for creating photoemission. What is the ratio of the work functions of the two emitters?
A.$1:2$
B.$2:1$
C.$1:4$
D.$4:1$

Answer 1

Hint: Work function is defined as the minimum amount of energy required to pull out the free electrons from the metal surface. Free electrons are defined as the electrons that are loosely bonded to the atom. These free electrons are in a metal surface weakly attracted by the nucleus. These free electrons can move from one atom to the other. But if minimum energy is applied to these free electrons they will jump from one atom to another. This minimum energy required to pull the electrons out from the metal surface is called a work function.

Complete answer:
The formula for work function is given as,
\[\phi = \dfrac{{hc}}{\lambda }\]
Here, \[\phi \]is the work function.
\[h\] is Planck’s constant.
\[c\] is the velocity of light.
\[\lambda \] is the wavelength of the light.
Let the work function of the first emitter be \[{\phi _1}\], and the work function of the second emitter be \[{\phi _2}\].
If we write the formula for both the emitters we get
\[{\phi _1} = \dfrac{{hc}}{{{\lambda _1}}}\] and \[{\phi _2} = \dfrac{{hc}}{{{\lambda _2}}}\] (Since \[h\] and \[c\] are constants.)
Therefore if we take the ratio between these two we get
\[\dfrac{{{\phi _1}}}{{{\phi _2}}} = \dfrac{{{\lambda _2}}}{{{\lambda _1}}}\]
Given that \[{\lambda _1} = 300nm\] and \[{\lambda _2} = 600nm\]. Therefore substituting this we get,
\[\dfrac{{{\lambda _2}}}{{{\lambda _1}}} = \dfrac{{600nm}}{{300nm}} = 2:1\]

Therefore the correct option is B.

Note:
The typical work function will vary from \[2eV\] to \[6eV\]. This value of work function usually depends on the nature and the purity of the metal surface. The nature of metal means how many electrons are present on the metal surface. The purity of the surface means if the metal is pure then the work function will be different.