Question

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5 \times {10^{ - 12}}m$ , the minimum electron energy required is close to
(A) $100keV$
(B) $500keV$
(C) $25keV$
(D) $1keV$

Answer 1

From the formula for the wavelength of electrons in terms of the Planck’s constant and the momentum, we can find the momentum as the wavelength is given as the same as the resolution. Now using the momentum we can find the energy as $K.E. = \dfrac{{{p^2}}}{{2m}}$ .

Formula used: In this solution we will be using the following formula,
$\lambda = \dfrac{h}{p}$
where $\lambda$ is the wavelength and $p$ is the momentum of the electron and
$h$ is the Planck’s constant.
$K.E. = \dfrac{{{p^2}}}{{2m}}$
where $K.E.$ is the energy of the electrons and $m$ is the mass.

Complete step by step solution:
In the question we are given that the wavelength of the electrons will be of the same order as the resolution of the electron microscope. So we are given the resolution to be $7.5 \times {10^{ - 12}}m$ . So this is the wavelength of the electrons. Hence we have,
$\lambda = 7.5 \times {10^{ - 12}}m$
The wavelength of an electron is given by the formula,
$\lambda = \dfrac{h}{p}$
Here we can take the $\lambda$ to the RHS and the $p$ to the LHS and get,
$p = \dfrac{h}{\lambda }$
The value of the Planck’s constant is $h = 6.6 \times {10^{ - 34}}kg{m^2}/s$
So substituting these values we get,
$p = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{7.5 \times {{10}^{ - 12}}}}$
Therefore, on calculating we have,
$p = 8.8 \times {10^{ - 23}}kgm/s$
Now the energy of the electrons are given by,
$K.E. = \dfrac{{{p^2}}}{{2m}}$
The mass of electrons is,
$m = 9.1 \times {10^{ - 31}}kg$
So substituting the values we get,
$K.E. = \dfrac{{{{\left( {8.8 \times {{10}^{ - 23}}} \right)}^2}}}{{2 \times \left( {9.1 \times {{10}^{ - 31}}} \right)}}$
On calculating we have,
$K.E. = \dfrac{{7.744 \times {{10}^{ - 45}}}}{{1.82 \times {{10}^{ - 30}}}}$
So we get the energy of the electrons as,
$K.E. = 4.2 \times {10^{ - 15}}J$
Now since the answer is given in electron volt, so we get,
$K.E. = \dfrac{{4.2 \times {{10}^{ - 15}}}}{{1.6 \times {{10}^{ - 19}}}}eV$
Calculating we get,
$K.E. = 25593eV \simeq 25keV$
Hence the correct answer is option C.

Note:
In this solution we have used the energy of electrons formula as $K.E. = \dfrac{{{p^2}}}{{2m}}$ . This is derived as,
Kinetic energy is given by $K.E. = \dfrac{1}{2}m{v^2}$
Multiplying $m$ in the numerator and denominator we get,
$K.E. = \dfrac{1}{{2m}}{m^2}{v^2}$ . Now since $p = mv$ , so we can write,
$K.E. = \dfrac{1}{{2m}}{p^2} = \dfrac{{{p^2}}}{{2m}}$