Question

What amount of accelerating potential is needed to produce an electron beam with an effective wavelength of 0.09A.

Answer 1

(here V is the velocity of electron)
amount of accelerating potential=V=182.03×10²

Answer 2

Explanation:

We assume that momentum (p) and mass (m) are required to find the potential difference (V). Therefore,  

              $\frac{P^{2}}{2m}$ = eV = gain in kinetic energy

where,   P = $\sqrt{2meV}$

According to De Broglie's equation,

                          $\lambda = \frac{h}{P}$

or,            $\lambda = \frac{h}{\sqrt{2meV}}$

where,       m = mass of an electron = $9.1 \times 10^{-31}$ kg

                  e = charge on an electron = $1.6 \times 10^{-19}$ C

      $\lambda$ = wavelength = 0.09 A = $9 \times 10^{-10}$ (as 1 $A^{o} = 10^{-10} m$)

                 h = plank's conctant = $6.626 \times 10^{-34}$

Therefore, putting the given values into the above formula as follows.

                         $\lambda = \frac{h}{\sqrt{2meV}}$

                $9 \times 10^{-10} = \frac{6.626 \times 10^{-34}}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times V}$

                                      V = $0.0252 \times 10^{74}$ V

Therefore, we can conclude that amount of accelerating potential is $0.0252 \times 10^{74}$ V.