Question

The work function of a metal is 4.2eV. Find its threshold wavelength.
A. \[{\rm{4000}}\mathop {\rm{A}}\limits^{\rm{0}} \]
B. \[{\rm{3500}}\mathop {\rm{A}}\limits^{\rm{0}} \]
C. \[{\rm{2960}}\mathop {\rm{A}}\limits^{\rm{0}} \]
D. \[{\rm{2500}}\mathop {\rm{A}}\limits^{\rm{0}} \]

Answer 1

Hint:Work function is the minimum energy required to remove one electron from a metal surface. Work function can also be said as the characteristics of the surface and not the complete metal. As we know that the energy of the incident light or photons should be more than the work function of the metal for the photoemission to take place. The minimum energy under which the emission of photoelectrons takes place is known as the threshold energy.

Formula Used:
To find the threshold wavelength we have,
\[{\lambda _0} = \dfrac{{hc}}{{{W_0}}}\]
Where, h is Planck’s constant, c is speed of light and \[{W_0}\] is work function.

Complete step by step solution:
They have given the work function, using this data we need to find the threshold wavelength. The formula to find the threshold wavelength is,
\[{\lambda _0} = \dfrac{{hc}}{{{W_0}}}\]
In order to calculate the threshold frequency first, we need to convert the work function from eV to joules. For this, we need to multiply the value of \[1.6 \times {10^{ - 19}}J\] to the work function. That is,
\[{W_0} = 4.2 \times 1.6 \times {10^{ - 19}}\]
\[\Rightarrow {W_0} = 6.72 \times {10^{ - 19}}J\]

Now, substitute the value of Planck’s constant, speed of light and the work function, them the above equation will become,
\[{\lambda _0} = \dfrac{{6.625 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{6.72 \times {{10}^{ - 19}}}}\]
\[\Rightarrow {\lambda _0} = 2.957 \times {10^{ - 7}}m\]
\[\therefore {{\rm{\lambda }}_{\rm{0}}} \simeq {\rm{2960}}\mathop {\rm{A}}\limits^{\rm{0}} \]
Therefore, the threshold wavelength will be \[{\rm{2960}}\mathop {\rm{A}}\limits^{\rm{0}} \].

Hence, Option C is the correct answer.

Additional information: Photon is the smallest packet (quanta) of energy and by the particle nature of light, we can say that the light behaves as particles and this is confirmed by the photoelectric effect, but according to the wave nature of light, light behaves as waves and is confirmed by phenomena like reflection, and refraction, etc.

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 while solving this problem.