Answer:
206 pm
Explanation:
We are given that
Potential difference,[tex]\Delta V=24.5+A V[/tex]
Distance,d=[tex]4.5+B cm[/tex]
We have to determine the de Brogile wavelength of the electron.
A=11 and B=5
[tex]\Delta V=24.5+11=35.5 V[/tex]
[tex]d=4.5+5=9.5 cm[/tex]
Charge on electron,q =[tex]1.6\times 10^{-19} C[/tex]
Mass of electron=m=[tex]9.1\times 10^{-31} kg[/tex]
Speed of electron,[tex]v=\sqrt{\frac{2q\Delta V}{m}}[/tex]
Using the formula
[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 35.5}{9.1\times 10^{-31}}[/tex]
v=[tex]3.53\times 10^6 m/s[/tex]
de Brogile wavelength, [tex]\lambda=\frac{h}{mv}[/tex]
Where [tex]h=6.626\times 10^{-34}[/tex]
[tex]\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[/tex]
1 pm=[tex]10^{-12} m[/tex]
[tex]\lambda=206 pm[/tex]
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.
[tex]\Delta V[/tex] = (24.5 + A) volts
By putting the value of A we will calculate [tex]\Delta V[/tex] as follows.
[tex]\Delta V[/tex] = (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 = [tex]q \times \Delta V[/tex]
[tex]\frac{1}{2}mv^{2} = 35.5 \times 1.6 \times 10^{-19}[/tex]
[tex]\frac{1}{2} \times 9.1 \times 10^{-31} \times v^{2} = 5.68 \times 10^{-18} J[/tex]
v = 3533202.016 m/s
Also we know that,
[tex]\lambda = \frac{r}{mv}[/tex]
= [tex]\frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 3533202.016 m/s}[/tex]
= [tex]206.2 \times 10^{-12}[/tex]
or, = 206.2 pm
Thus, we can conclude that the de Broglie wavelength of the electron is 206.2 pm.
