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 wav

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.

in progress 0
Nem 5 years 2021-08-05T04:47:35+00:00 2 Answers 16 views 0

Answers ( )

  1. kimchi2
    0
    2021-08-05T04:48:35+00:00
    Reply

    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.

  2. hongcuc2
    0
    2021-08-05T04:48:57+00:00
    Reply

    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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )