Question

An electron is accelerated from rest by a potential difference of (24.5 A) V for a distance of (4.50 B) cm. Determine the de Broglie wavelength of the electron. Give your answer in picometers (pm) and with 3 significant figures.

Answer 1

Answer:

206 pm

Explanation:

We are given that

Potential difference,$\Delta V=24.5+A V$

Distance,d=$4.5+B cm$

We have to determine the de Brogile wavelength of the electron.

A=11 and B=5

$\Delta V=24.5+11=35.5 V$

$d=4.5+5=9.5 cm$

Charge on electron,q =$1.6\times 10^{-19} C$

Mass of electron=m=$9.1\times 10^{-31} kg$

Speed of electron,$v=\sqrt{\frac{2q\Delta V}{m}}$

Using the formula

$v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 35.5}{9.1\times 10^{-31}}$

v=$3.53\times 10^6 m/s$

de Brogile wavelength, $\lambda=\frac{h}{mv}$

Where $h=6.626\times 10^{-34}$

$\lambda=\frac{6.626\times 10^{-34}}{9.1\times 10^{-31}\times 3.53\times 10^6}=2.06\times 10^{-10}=206\times 10^{-12} m$

1 pm=$10^{-12} m$

$\lambda=206 pm$

Answer 2

Given question is incomplete. The complete question is as follows.

An electron is accelerated from rest by a potential difference of (24.5 + A) V for a distance of (4.50 + B) cm. Determine the de Broglie wavelength of the electron. Give your answer in picometers (pm) and with 3 significant figures.

A = 11

B = 5

Explanation:

Change in potential difference will be as follows.

     $\Delta V$ = (24.5 + A) volts

By putting the value of A we will calculate $\Delta V$ as follows.

      $\Delta V$ = (24.5 + A) volts

                     = (24.5 + 11) volts    

                     = 35.51 volts

and,    d = (4.50 + B) cm

             = (4.50 + 5) cm

             = 9.50 cm  

Now,   kinetic energy = $q \times \Delta V$

            $\frac{1}{2}mv^{2} = 35.5 \times 1.6 \times 10^{-19}$

  $\frac{1}{2} \times 9.1 \times 10^{-31} \times v^{2} = 5.68 \times 10^{-18} J$

                v = 3533202.016 m/s

Also we know that,

       $\lambda = \frac{r}{mv}$

                   = $\frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 3533202.016 m/s}$    

                   = $206.2 \times 10^{-12}$

or,                = 206.2 pm

Thus, we can conclude that the de Broglie wavelength of the electron is 206.2 pm.