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Question: The work function for a metal is 4 eV. To emit a photo electron of zero velocity from the surface of...

The work function for a metal is 4 eV. To emit a photo electron of zero velocity from the surface of the metal, the wavelength of incident light should be:
A. 2700A∘2700A^\circ
B. 1700A∘1700A^\circ
C. 5900A∘5900A^\circ
D. 3100A∘3100A^\circ

Explanation

Solution

Work function is defined for metals. It is defined as the minimum energy required to extract one electron from a metal. Work function can be said as the characteristics of the surface and not the complete metal. As we know The energy of the incident light or photons should be more than the work function of the metal for photoemission to take place.

Complete step by step answer:
When an electron is removed from a metal surface, it gains positive charge. Hence the negative charge of the electron and the positive charge of the metal gets attracted such that the electron turns back to the metal surface. To overcome this attraction, energy is supplied to the metal so that an electron can be escaped from the metal.
As we already discussed, only if the energy of the incident light is greater than the threshold energy of the metal surface, electrons can be emitted.
So we can write as hν>hνoh\nu > h{\nu _o}
hν−hνo=K.E  ofelectronemittedh\nu - h{\nu _o} = K.E\;{{of electron emitted}}, kinetic energy
hν−hνo=K.E  h\nu - h{\nu _o} = K.E\;
Where h is the Planck's constant, ν\nu is the frequency of incident light and νo{\nu _o}is the threshold frequency.
hνoh{\nu _o} is the work function which is 4 eV
In the question it is given that a photoelectron of zero velocity is emitted. So we can say that the kinetic energy is Zero as K.E=12mv2K.E = \dfrac{1}{2}m{v^2}
So, hν=hνoh\nu = h{\nu _o}
hν=4eVh\nu = 4eV
Frequency of incident light, ν=4h\nu = \dfrac{4}{h}
We know that Planck's constant is 6.626×10−346.626 \times {10^{ - 34}}
ν=4eV6.626×10−34\nu = \dfrac{{4eV}}{{6.626 \times {{10}^{ - 34}}}}
We know that frequency,ν=cλ\nu = \dfrac{c}{\lambda }, c is the speed of light and λ\lambda is the wavelength of the incident light.
λ=cν\lambda = \dfrac{c}{\nu }
λ=3×1084eV6.626×10−34\lambda = \dfrac{{3 \times {{10}^8}}}{{\dfrac{{4eV}}{{6.626 \times {{10}^{ - 34}}}}}}
λ=3×108×6.626×10−344eV\lambda = \dfrac{{3 \times {{10}^8} \times 6.626 \times {{10}^{ - 34}}}}{{4eV}}
λ=3×108×6.626×10−344×1.6×10−19=3100A∘\lambda = \dfrac{{3 \times {{10}^8} \times 6.626 \times {{10}^{ - 34}}}}{{4 \times 1.6 \times {{10}^{ - 19}}}} = 3100A^\circ

So, the correct answer is Option D.

Note: c is the speed of light with value 3×1083 \times {10^8} and the value of 1eV=1.6×10−19V1eV = 1.6 \times {10^{ - 19}}V
In metals, there are many free electrons that are loosely bound to the atom. Typical value of work function varies from 2ev to 6eV