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Question: A metal is irradiated with light of wavelength\[600nm\]. Given that the work function of the metal i...

A metal is irradiated with light of wavelength600nm600nm. Given that the work function of the metal is 1.0eV1.0eV. The de Broglie wavelength of the ejected electron is close to:
A. 6.6×10−7m6.6 \times {10^{ - 7}}m
B. 8.9×10−11m8.9 \times {10^{ - 11}}m
C. 1.3×10−9m1.3 \times {10^{ - 9}}m
D. 6.6×10−13m6.6 \times {10^{ - 13}}m

Explanation

Solution

To solve this question we will consider the concepts of incident light and use the formula of energy of an incident light. Also, we will make use of the Kinetic energy of a photoelectron by de-Broglie's justification to Bohr’s assumption.

Complete step by step answer: Let’s first note down all the given quantities:
λ\lambda (De Broglie’s wavelength) =600nm=600×10−9m = 600nm = 600 \times {10^{ - 9}}m
Work function, w.f. = 1.0eV=1.0×10−19×1.6J1.0eV = 1.0 \times {10^{ - 19}} \times 1.6J
Also, we know that work function w.f. = hv0h{v_0}
Thus, we know that total energy is equal to the summation of work function and kinetic energy. Therefore, it is represented as hv=hv0+K.E.hv = h{v_0} + K.E.
From de Broglie’s justification to Bohr’s relationship we know that,
K.E=p22mK.E = \dfrac{{{p^2}}}{{2m}}= h22mλ2\dfrac{{{h^2}}}{{2m{\lambda ^2}}}
Here, we also know that de Broglie’s wavelength λ\lambda =hp\dfrac{h}{p}
As we know that the energy of an incident light is
E=hcλE = \dfrac{{hc}}{\lambda }
where E is the photon energy, h is the Planck’s constant, λ\lambda is the wavelength of a photoelectron and c is the speed of light in vacuum.
Then, also we know from Planck’s Einstein relation that
E=hϑE = h\vartheta
Where h is the Planck’s constant and ϑ\vartheta is the photon’s frequency.
Thus, E = hv=hcλhv = \dfrac{{hc}}{\lambda }
Let’s place all the values in this equation:
E= 6.625×10−34×3×108600×10−9\dfrac{{6.625 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{600 \times {{10}^{ - 9}}}} = 3.313×10−9J3.313 \times {10^{ - 9}}J
Here, we will substitute the formula of kinetic energy
K.E=hv−hv0K.E = hv - h{v_0} (Since,E=hvE = hv and W=hv0W = h{v_0} )
K.E=E−WK.E = E - W = (3.313-1.16)×10−9J \times {10^{ - 9}}J
As we know that, h22mλ2=1.713×10−19J\dfrac{{{h^2}}}{{2m{\lambda ^2}}} = 1.713 \times {10^{ - 19}}J
Thus, λ2=(6.626×10−34)22(9.1×10−31)(1.713×10−19){\lambda ^2} = \dfrac{{{{(6.626 \times {{10}^{ - 34}})}^2}}}{{2(9.1 \times {{10}^{ - 31}})(1.713 \times {{10}^{ - 19}})}}
So, λ=1.28×10−9m\lambda = 1.28 \times {10^{ - 9}}m
Therefore, de Broglie’s wavelength of the ejected electron is approximately equal to 1.3×10−9m1.3 \times {10^{ - 9}}m.

So, the correct answer is “Option C”.

Note: We must know that De Broglie did not have any actual experimental evidence for his conjecture. Clinton J. Davisson and Lester H. Germer were shooting electron particles in a nickel crystal. What they saw was electron diffraction similar to wave diffraction against crystals (x-rays). George P. Thomson also fired electrons into thin metal foil, giving him the same results as Davisson and Germer.