Question

An electron of energy $150\;{\rm{eV}}$ has wavelength of ${10^{ - 10}}\;{\rm{m}}$. The wavelength of an $0.60\;{\rm{keV}}$ electron is:

Answer 1

The above problem can be resolved using the concept and fundamentals of the energy and the wavelength associated with that energy. The mathematical relation for the energy is related directly with the square of the wavelength. In the given problem, the value of one energy is given along with the value of wavelength corresponding to it and also with the magnitude of another energy associated with the electron. Then by applying the substitution, the desired value of wavelength can be calculated.

Complete step by step answer:
Given:
The energy of an electron is, ${E_1} = 150\;{\rm{eV}}$.
The wavelength is, ${\lambda _1} = {10^{ - 10}}\;{\rm{m}}$.
The another value of energy of electron is,

{E_2} = 60\;{\rm{keV}}\\\ {E_2} = 60\;{\rm{keV}} \times \dfrac{{1000\;{\rm{eV}}}}{{1\;{\rm{keV}}}}\\\ {E_2} = 60 \times {10^3}\;{\rm{eV}} \end{array}$$ The expression for the energy of electron for the unknown value of wavelength is given as, $$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \sqrt {\dfrac{{{E_2}}}{{{E_1}}}} $$ Here, $${E_2}$$ is the value of energy corresponding to the wavelength $${\lambda _2}$$. Solve by substituting the values in the above equation as, $$\begin{array}{c} \dfrac{{{{10}^{ - 10}}\;{\rm{m}}}}{{{\lambda _2}}} = \times \sqrt {\dfrac{{0.6\;{\rm{keV}} \times \dfrac{{1000\;{\rm{eV}}}}{{1\;{\rm{keV}}}}}}{{150\;{\rm{eV}}}}} \\\ {\lambda _2} = {10^{ - 10}} \times 2\;\;{\rm{m}}\\\ {\lambda _2} = 0.5 \times {10^{ - 10}}\;{\rm{m}} \end{array}$$ Therefore, the wavelength of an $$0.60\;{\rm{keV}}$$ electron is $$0.5 \times {10^{ - 10}}\;{\rm{m}}$$. **Note:** Try to understand the fundamentals behind the energy released by electrons and the wavelength associated with that value of energy. When electrons are ejected, some magnitude of energy is released, and by applying the mathematical formula, there is a wavelength associated with this magnitude of energy. Moreover, the concept and application of the photoelectric effect are also needed to consider, as this accounts for direct linkage with the applied concept in the problem.