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Question: The work function of a metal is Light of wavelength that is incident on this metal surface. The velo...

The work function of a metal is Light of wavelength that is incident on this metal surface. The velocity of emitted photoelectrons will be:

Explanation

Solution

Under the right circumstances light can be used to push electrons, freeing them from the surface of a solid. This process is called the photoelectric effect. The velocity & the kinetic energy and the energy of the liberated photo electron are dependent on each other. In order to find the value of the velocity of the electron particle, just compare the two formulas which describe the value of energy of the electron particle.

Complete step by step solution:
Here we need to find the value of the velocity of the electron particle. Let’s define the terms given in the question:
Work function, W=1.0eVW = 1.0eV
Wavelength of light, λ=3000Ao\lambda = 3000\mathop {\rm A}\limits^o
From the Einstein’s photoelectric theory we know, the energy of the photoelectron,
E=W+K.EE = W + K.E……………………… (1)
Here, W=1.0ev=1×1.6×10−19JW = 1.0ev = 1 \times 1.6 \times {10^{ - 19}}J
K.EK.E is the kinetic energy.
Let’s find out the value of the energy of photoelectron,
E=hcλE = \dfrac{{hc}}{\lambda }……………… (2)
Where, hh is the Planck's constant and it is given by,
h=6.6×10−34m2kgs−1h = 6.6 \times {10^{ - 34}}{m^2}kg{s^{ - 1}}
cc is the speed of light, and is given by,
c=3×108ms−1c = 3 \times {10^8}m{s^{ - 1}}
λ\lambda is the wavelength of the light,
λ=3000Ao=3000×10−10m\lambda = 3000\mathop {\rm A}\limits^o = 3000 \times {10^{ - 10}}m
We are applying all the known values to this equation (2)
We get, E=6.6×10−34×3×1083000×10−10E = \dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3000 \times {{10}^{ - 10}}}}
⇒E=6.6×10−34×108+7\Rightarrow E = 6.6 \times {10^{ - 34}} \times {10^{8 + 7}}
⇒E=6.6×10−19J\Rightarrow E = 6.6 \times {10^{ - 19}}J
We are converting this value from JJ into eVeV
⇒E=6.6×10−191.6×10−19eV\Rightarrow E = \dfrac{{6.6 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}}eV
⇒E=4.125eV\Rightarrow E = 4.125eV
Now we know the value of energy of the photoelectron, E=4.125eVE = 4.125eV
Applying this value of energy of photoelectron in the equation (2)
We get, E=W+K.EE = W + K.E
Apply the value of energy of the photoelectron, E=4.125eVE = 4.125eV and the value of work function, W=1.0eVW = 1.0eV
We get, E=4.125eV=1eV+K.EE = 4.125eV = 1eV + K.E
⇒K.E=4.125−1\Rightarrow K.E = 4.125 - 1
⇒K.E=3.125eV\Rightarrow K.E = 3.125eV
⇒K.E=3.125×1.6×10−19J\Rightarrow K.E = 3.125 \times 1.6 \times {10^{ - 19}}J
Now we have the value of the kinetic energy of the photoelectron. We can calculate the value of the velocity from the equation for the kinetic energy.
It is known that the kinetic energy of the particle is given by,
K.E=12mv2K.E = \dfrac{1}{2}m{v^2}……………….. (3)
Here, mmis the mass of the photoelectron and is given by, m=9.1×10−31kgm = 9.1 \times {10^{ - 31}}kg
Applying all the values to this equation (3), we get,
⇒K.E=3.125×1.6×10−19J=12×9.1×10−31v2\Rightarrow K.E = 3.125 \times 1.6 \times {10^{ - 19}}J = \dfrac{1}{2} \times 9.1 \times {10^{ - 31}}{v^2}
⇒v2=2×3.125×1.6×10−199.1×10−31\Rightarrow {v^2} = \dfrac{{2 \times 3.125 \times 1.6 \times {{10}^{ - 19}}}}{{9.1 \times {{10}^{ - 31}}}}
⇒v=2×3.125×1.6×10−199.1×10−31\Rightarrow v = \sqrt {\dfrac{{2 \times 3.125 \times 1.6 \times {{10}^{ - 19}}}}{{9.1 \times {{10}^{ - 31}}}}}
⇒v=1×106ms−1\Rightarrow v = 1 \times {10^6}m{s^{ - 1}}

The final answer is the Option (D), 1×106ms−11 \times {10^6}m{s^{ - 1}}.

Note: Einstein's explanation of the photoelectric effect was very simple. He assumed that the kinetic energy of the ejected electron was equal to the energy of the incident photon minus the energy required to remove the electron from the material, which is called the work function.