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Question: Photoelectrons are liberated by the ultraviolet light of wavelength \[3000\mathop {\text{A}}\limits^...

Photoelectrons are liberated by the ultraviolet light of wavelength 3000A∘3000\mathop {\text{A}}\limits^\circ from a metallic surface for which the photoelectric threshold is 4000A∘4000\mathop {\text{A}}\limits^\circ . The de-Broglie wavelength of electrons emitted with maximum kinetic energy is:
A) 1.2nm
B) 3.215 nm
C) 7.28 A∘\mathop {\text{A}}\limits^\circ
D) 1.65 A∘\mathop {\text{A}}\limits^\circ

Explanation

Solution

Using the given wavelength of Photoelectrons calculate the energy of the photon. Similarly, using the threshold wavelength calculate the threshold energy. Using the calculated photon energy and threshold energy calculate the maximum kinetic energy of the electron. Using the kinetic energy calculate the de-Broglie wavelength of the electron.

Formula Used: The equation for the energy of a photon is:
E = hcλE{\text{ = }}\dfrac{{hc}}{\lambda }
The threshold energy equation is :
E∘ = hcλ∘E^\circ {\text{ = }}\dfrac{{hc}}{{\lambda ^\circ }}
The equation for Kinetic energy is:
K.E.=E−E∘K.E. = E - E^\circ
K.E.=12mv2K.E. = \dfrac{1}{2}m{v^2}
De-Broglie wavelength
λ = hmv\lambda {\text{ = }}\dfrac{h}{{mv}}

Complete step by step answer:
To calculate the maximum kinetic energy we need to calculate photon energy and threshold energy.
Calculate the energy of photon as follows:
E = hcλE{\text{ = }}\dfrac{{hc}}{\lambda }
Here,
E E{\text{ }}= energy of the photon
hh= Planck’s constant = 6.626×10−34Js6.626 \times {10^{ - 34}}{\text{Js}}
cc= speed of light in vacuum = 3.0×108m/s{\text{3}}{\text{.0}} \times {\text{1}}{{\text{0}}^8}{\text{m/s}}
λ\lambda = wavelength = 3000A∘3000\mathop {\text{A}}\limits^\circ = 3000×10−10m{\text{3000}} \times {\text{1}}{{\text{0}}^{ - 10}}{\text{m}}
( As 1A∘\mathop {\text{A}}\limits^\circ =10−10m{\text{1}}{{\text{0}}^{ - 10}}{\text{m}})
Now, We can substitute 3000×10−10m{\text{3000}} \times {\text{1}}{{\text{0}}^{ - 10}}{\text{m}} for the wavelength of a photoelectron, 6.626×10−34Js6.626 \times {10^{ - 34}}{\text{Js}} for Planck’s constant, 3.0×108m/s{\text{3}}{\text{.0}} \times {\text{1}}{{\text{0}}^8}{\text{m/s}} for speed of light and calculate the energy of the photon.
E = 6.626×10−34Js×3.0×108m/s3000×10−10m=6.626×10−19JE{\text{ = }}\dfrac{{6.626 \times {{10}^{ - 34}}{\text{Js}} \times {\text{3}}{\text{.0}} \times {\text{1}}{{\text{0}}^8}{\text{m/s}}}}{{{\text{3000}} \times {\text{1}}{{\text{0}}^{ - 10}}{\text{m}}}} = 6.626 \times {10^{ - 19}}{\text{J}}
Similarly, we can calculate the threshold energy using the threshold wavelength as follows:
E∘ = hcλ∘E^\circ {\text{ = }}\dfrac{{hc}}{{\lambda ^\circ }}
λ∘\lambda ^\circ = threshold wavelength = 4000A∘4000\mathop {\text{A}}\limits^\circ =4000×10−10m{\text{4000}} \times {\text{1}}{{\text{0}}^{ - 10}}{\text{m}}
E = 6.626×10−34Js×3.0×108m/s4000×10−10m=4.9695×10−19JE{\text{ = }}\dfrac{{6.626 \times {{10}^{ - 34}}{\text{Js}} \times {\text{3}}{\text{.0}} \times {\text{1}}{{\text{0}}^8}{\text{m/s}}}}{{{\text{4000}} \times {\text{1}}{{\text{0}}^{ - 10}}{\text{m}}}} = 4.9695 \times {10^{ - 19}}{\text{J}}
Now, we have the energy of photon and threshold energy so we can calculate the kinetic energy as follows:
K.E.=E−E∘K.E. = E - E^\circ
Substitute 6.626×10−19J6.626 \times {10^{ - 19}}{\text{J}} for the energy of the photon and 4.9695×10−19J4.9695 \times {10^{ - 19}}{\text{J}} for threshold energy.
K.E.=6.626×10−19J−4.9695×10−19J=1.6565×10−19JK.E. = 6.626 \times {10^{ - 19}}{\text{J}} - 4.9695 \times {10^{ - 19}}{\text{J}} = 1.6565 \times {10^{ - 19}}{\text{J}}
Now, using the kinetic energy calculate the de-Broglie wavelength of an electron as follows:
K.E.=12mv2K.E. = \dfrac{1}{2}m{v^2}
Where,
mm = mass of electron = 9.1×10−31kg9.1 \times {10^{ - 31}}{\text{kg}}
vv = velocity of electron
mv2=2×K.E.m{v^2} = 2 \times K.E.
Multiply both the side of the equation by m.
So, m2v2=2×K.E.×m{m^2}{v^2} = 2 \times K.E. \times m
m2v2=2×1.6565×10−19J×9.1×10−31kg{m^2}{v^2} = 2 \times 1.6565 \times {10^{ - 19}}{\text{J}} \times 9.1 \times {10^{ - 31}}{\text{kg}}
m2v2= 3.01483×10−49{m^2}{v^2} = {\text{ 3}}{\text{.01483}} \times {\text{1}}{{\text{0}}^{ - 49}}
mv= 5.49×10−25mv = {\text{ 5}}{\text{.49}} \times {\text{1}}{{\text{0}}^{ - 25}}
Now, substitute  5.49×10−25{\text{ 5}}{\text{.49}} \times {\text{1}}{{\text{0}}^{ - 25}} in the de-Broglie wavelength equation.
λ = hmv\lambda {\text{ = }}\dfrac{h}{{mv}}
λ = 6.626×10−34 5.49×10−25=1.2×10−9m\lambda {\text{ = }}\dfrac{{6.626 \times {{10}^{ - 34}}}}{{{\text{ 5}}{\text{.49}} \times {\text{1}}{{\text{0}}^{ - 25}}}} = 1.2 \times {10^{ - 9}}{\text{m}}
1nm = 1.0×10−9m1.0 \times {10^{ - 9}}{\text{m}}
λ = 1.2 nm\lambda {\text{ = }}1.2{\text{ nm}}
Thus, the de-Broglie wavelength of electrons emitted with maximum kinetic energy is 1.2 nm.

Hence, the correct option is (A) 1.2nm.

Note: The energy of a photon is the sum of the kinetic energy of the electron and threshold energy. Threshold wavelength is the maximum wavelength incident radiation necessary to emit a photon. de-Broglie pointed out that electrons have dual nature. It can behave as a wave as well as a particle.