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Question: Light of wavelength \(500nm\) is incident on a metal with work function \[2.28eV\]. The de Broglie...

Light of wavelength 500nm500nm is incident on a metal with work function 2.28eV2.28eV.
The de Broglie wavelength of the emitted electron is-
(A) ≤2.8×10−12m\le 2.8\times {{10}^{^{-12}}}m
(B) <2.8×10−10m<2.8\times {{10}^{^{-10}}}m
(C) <2.8×10−9m<2.8\times {{10}^{^{-9}}}m
(D) ≥2.8×10−9m\ge 2.8\times {{10}^{^{-9}}}m

Explanation

Solution

Hint The De Broglie wavelength is a wavelength manifested in all objects in quantum mechanics, according to wave-particle duality, which defines the probability density of locating the object at a given point in the space of the configuration. A particle's de Broglie wavelength is inversely proportional to its momentum.

Complete step by step answer
Albert Einstein proposed an equation in 1905, the annus mirabilis (miracle year) of physics, to describe this effect. Einstein argued that light was a wave in the form of a packet of energy or a quantum of energy which interacts with matter. This quantum of radiation was a photon and is called Einstein's photoelectric equation.
Thus, Einstein’s photoelectric equation gives us-
E=w0+eVE={{w}_{0}}+eV
The vision of light from Einstein was both beautiful and groundbreaking. He suggested a peculiar but powerful radiation model. The light was made up of very tiny particles. It was not matter but pure energy that these particles were. He called each one of these a radiation quantum. Light must then be made up of these qantas, or energy or quantum energy bundles. We call them photons and they bear our light source's momentum and energy.
As energy of photon, λ≥2.8×10−9m\lambda \ge 2.8\times {{10}^{-9}}m

Since, ν=cλ\nu =\dfrac{c}{\lambda }
⇒E=hcλ\Rightarrow E=\dfrac{hc}{\lambda }
Where,
h=6.6×10−34h=6.6\times {{10}^{-34}}
c=3×108c=3\times {{10}^{8}}
V≤0.2V\le 0.2
⇒E=6.626×10−34×3×108500×10−9\Rightarrow E=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{500\times {{10}^{-9}}}
=0.0397×10−21J=0.0397\times {{10}^{-21}}J
Now, hcλ=w0+eV\dfrac{hc}{\lambda }={{w}_{0}}+eV
So we can write, ⇒0.0397×10−211.6×10−19=0.0248×102eV\Rightarrow \dfrac{0.0397\times {{10}^{-21}}}{1.6\times {{10}^{-19}}}=0.0248\times {{10}^{2}}eV
=2.48eV=2.48eV
According to Einstein photoelectric emission,
We have,
KEmax⁡imum=E−WK{{E}_{\max imum}}=E-W
⇒KEmax⁡imum=2.48−2.38=0.2eV\Rightarrow K{{E}_{\max imum}}=2.48-2.38=0.2eV
And, the question asks de-broglie wavelength of the emitted electron. And for making wavelength minimum the denominator should be maximum-
λmin⁡imum=12.27KEmax⁡imumeV{{\lambda }_{\min imum}}=\dfrac{12.27}{\sqrt{K{{E}_{\max imum}}eV}}
⇒λmin⁡imum=12.270.2eV\Rightarrow {{\lambda }_{\min imum}}=\dfrac{12.27}{\sqrt{0.2eV}}
Thus, after calculating this the minimum wavelength is found to be
⇒λmin⁡imum=2.7436×10−9m\Rightarrow {{\lambda }_{\min imum}}=2.7436\times {{10}^{-9}}m
and, λ≥λmin⁡imumand, \, \lambda \ge {{\lambda }_{\min imum}}
So, λ≥2.8×10−9m\lambda \ge 2.8\times {{10}^{-9}}m

Therefore, the correct answer is option D.

Note: The maximum kinetic energy of the photoelectrons emitted is directly proportional to the incident radiation frequency. This means that the photoelectron 's overall kinetic energy depends only on the strength of the incident light.