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Question: Photoelectric work function of a metal is 1 \(eV\). Light of wavelength \(\lambda = 3000{A^0}\) fall...

Photoelectric work function of a metal is 1 eVeV. Light of wavelength λ=3000A0\lambda = 3000{A^0} falls on it. The photoelectrons come out with velocity:
A) 10ms−110m{s^{ - 1}}
B) 103ms−1{10^3}m{s^{ - 1}}
C) 104ms−1{10^4}m{s^{ - 1}}
D) 106ms−1{10^6}m{s^{ - 1}}

Explanation

Solution

The phenomenon in which the electrons or charged particles absorb electromagnetic radiation and are ejected from the surface of the material is known as photoelectric effect. It can also be defined as the ejection of electrons from a surface when light is incident on it.

Complete step by step answer:
According to photoelectric effect, the energy of incident photons is equal to the energy required to remove an electron from the surface and kinetic energy of the emitted electron. When the energy of incident photons on metals is just sufficient to escape an electron or it is greater than that binding energy (work function) photo electrons are emitted.
hν=W+K.E.h\nu = W + K.E.
where hh is Planck's constant=6.62×10−34 = 6.62 \times {10^{ - 34}}
ν\nu is frequency of incident photon
WW is the work function=1eV=1.6×10−19J = 1eV = 1.6 \times {10^{ - 19}}J
K.E.K.E. is kinetic energy of the electron

Step II:
Now applying Einstein photoelectric equation for maximum kinetic energy of photoelectrons
(K.E)max⁡=hν−W−−−−−−(i){(K.E)_{\max }} = h\nu - W - - - - - - (i)
ν\nu is the frequency and its formula is ν=cλ\nu = \dfrac{c}{\lambda }
where λ\lambda is the wavelength of the incident light=3000A∘=3000×10−10m = 3000{A^ \circ } = 3000 \times {10^{ - 10}}m
cc is the speed of light
And formula for kinetic energy is=12mv2 = \dfrac{1}{2}m{v^2}
wheremm is the mass of the ejected electron=9.1×10−31kg = 9.1 \times {10^{ - 31}}kg

Step III:
Substituting the values in equation (i) and solving for ‘v’
12mv2=hcλ−W\dfrac{1}{2}m{v^2} = \dfrac{{hc}}{\lambda } - W
v=2(hcλ−W)mv = \sqrt {\dfrac{{2(\dfrac{{hc}}{\lambda } - W)}}{m}}
ν=2[6.62×10−34×3×108)3000×10−10−(1×1.6×10−10)]9.1×10−31\nu = \sqrt {\dfrac{{2[\dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8})}}{{3000 \times {{10}^{ - 10}}}} - (1 \times 1.6 \times {{10}^{ - 10}})]}}{{9.1 \times {{10}^{ - 31}}}}}
ν=39.72×10−263000×10−10−3.2×10−109.1×10−31\nu = \sqrt {\dfrac{{\dfrac{{39.72 \times {{10}^{ - 26}}}}{{3000 \times {{10}^{ - 10}}}} - 3.2 \times {{10}^{ - 10}}}}{{9.1 \times {{10}^{ - 31}}}}}
ν=13.2×10−13−3.2×10−109.1×10−31\nu = \sqrt {\dfrac{{13.2 \times {{10}^{ - 13}} - 3.2 \times {{10}^{ - 10}}}}{{9.1 \times {{10}^{ - 31}}}}}
ν=1.04×106m/s\nu = 1.04 \times {10^6}m/s
Or ν∼106m/s\nu \sim {10^6}m/s
Step IV:
The velocity by which photoelectrons come out is vmax⁡=106m/sec⁡{v_{\max }} = {10^6}m/\sec
Option D is the right answer.

Note: Sometimes the term work function can be mixed with threshold frequency. But they both are different. Work function is the minimum energy required to move an electron from to infinity from the surface of the metal. Work function depends on the structure and composition of the surface. But threshold frequency is the minimum frequency of light that is required to provide energy to the work function.