Question

A beam of electrons is accelerated by a potential difference of $10000volts$. The wavelength of wave associated with it will be:
A.$0.0123\mathop {\text{A}}\limits^ \circ $
B.$1.23\mathop {\text{A}}\limits^ \circ $
C.$0.123\mathop {\text{A}}\limits^ \circ $
D.None

Answer 1

Hint:We know that the mass of an electron is $9.1 \times {10^{ - 31}}Kg$, the charge of an electron $1.6 \times {10^{ - 19}}C$. And the relation between the wavelength and potential difference is as $\lambda = \dfrac{h}{{\sqrt {2meV} }}$, where $h$ is planck's constant, $m$ is the mass of an electron, $e$ is charge of an electron and $V$ is the potential difference of the accelerated electron beam. Now apply this equation and you will get the answer.

Complete step by step answer:
Let us first talk about mass of an electron, charge of an electron and the potential difference.
Mass of an electron is defined as the mass of one electron of an atom which is $9.1 \times {10^{ - 31}}Kg$.
Charge of an electron: It is defined as the charge of an electron which is as $1.6 \times {10^{ - 19}}C$. And we know that electrons are negatively charged particles of an atom.
Potential difference: It is defined as the work involved in the transfer of a unit quantity of electricity from one point to the other. It is expressed in volts.
Now wavelength is defined as the distance between the identical points i.e. between adjacent crests or adjacent troughs, of a waveform signal. It is expressed by $\lambda $ and measured in $m$.
Now let us talk about when an electron beam is accelerated by a potential difference then we can calculate the wavelength of an electron if we know the potential difference of an accelerated electron beam (the rays of electrons in which electrons are aligned). The relation between wavelength and potential difference of the accelerated electron beam is as: $\lambda = \dfrac{h}{{\sqrt {2meV} }}$, where $h$ is planck's constant, $m$ is the mass of an electron, $e$is charge of an electron and $V$ is the potential difference of the accelerated electron beam.. This relation was given by de-Broglie so the wavelength calculated by this relation is known as de-Broglie wavelength.
Now here in the question we are given with the potential difference of an accelerated electron beam as $10000volts = {10^4}volts$and we know the values of other terms which are as: $h = 6.6 \times {10^{ - 34}},m = 9.1 \times {10^{ - 31}},e = 1.6 \times {10^{ - 19}}$. Now put these values in de-Broglie equation and you will get the value of wavelength as: $\lambda = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}} \times {{10}^4}} }} = 0.123\mathop {\text{A}}\limits^ \circ $.

Hence option C is correct.

Note:




Along with electrons there are two more particles in an atom which are known as protons and neutrons. And electrons are negatively charged particles, protons are positively charged particles and neutrons are neutral (having no charge) particles.