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Question: The energy of electron with de - Broglie wavelength of \({10^{ - 10}}\)metre, is: (A) 13.6 eV (B...

The energy of electron with de - Broglie wavelength of 10−10{10^{ - 10}}metre, is:
(A) 13.6 eV
(B) 12.27eV
(C) 1.227eV
(D) 150.5eV

Explanation

Solution

We are already provided with the wavelength. The De Broglie hypothesis can be used to solve this question. To calculate the amount of energy required:
Using De- Broglie hypothesis
 λ=h2×KE×me   \ {{\lambda = }}\dfrac{{{h}}}{{\sqrt {{{2 \times KE \times }}{{{m}}_{{e}}}} }} \\\ \\\ \
Here, λ\lambda is the De- Broglie wavelength,
h{{h}} is Planck’s constant,
KE{{KE}} is the energy of electron,
and me{{{m}}_{{e}}} is the mass of the electron.

Complete step by step solution:
We are already provided with the wavelength.
By simply applying the de Broglie hypothesis,
Also, we know the value of Planck’s constant h=9.1×10−31Js{{h = 9}}{{.1 \times 1}}{{{0}}^{{{ - 31}}}}{{Js}}
And mass of electron me=9.1×10−31Js{{{m}}_e}{{ = 9}}.{{1}} \times {{1}}{{{0}}^{{{ - 31}}}}{{Js}}
So, KE=h22λ2me{{KE = }}\dfrac{{{{{h}}^{{2}}}}}{{{{2}}{\lambda ^{{2}}}{{{m}}_{{e}}}}}
KE=(6.626×10−34)22(10−10)2×9.1×10−31{{KE = }}\dfrac{{{{\left( {6.626 \times {{10}^{ - 34}}} \right)}^2}}}{{{{2}}{{\left( {{{10}^{ - 10}}} \right)}^{{2}}} \times 9.1 \times {{10}^{ - 31}}}}
To convert to eV:
KE=(6.626×10−34)22(10−10)2×9.1×10−31×1.6×10−19{{KE = }}\dfrac{{{{\left( {6.626 \times {{10}^{ - 34}}} \right)}^2}}}{{{{2}}{{\left( {{{10}^{ - 10}}} \right)}^{{2}}} \times 9.1 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}}}}
KE = 150.7688 eV
So, we need to see from the above options, and select the approximate value.
Thus, the correct answer is option D.

Additional information:
Louis de Broglie was an eminent French physicist. He gained worldwide acclaim for his ground-breaking work on quantum theory. In his 1924 thesis, he discovered the wave nature of electrons and suggested that all matter have wave properties. This was known as the De Broglie hypothesis.
De Broglie concluded that most particles are too heavy to observe their wave properties. When the mass of an object is very small, however, the wave properties can be detected experimentally. De Broglie predicted that the mass of an electron was small enough to exhibit the properties of both particles and waves.

Note:
The common mistake that can be done here is in taking the values of Planck's constant and mass of the electron as these values are fixed and most of the time mentioned in the questions.
Also, never forget to convert your answer into the units as required by the options given for the questions. Here the Kinetic energy has to be converted in eV to get the correct answer.