What accelerating potential difference would be required to give each of the following particles a de broglie wavelength of 2.0nm?
a) a proton
b)an alpha particle
The answer is
a)2.0x^-4 V
b)2.57x10^-5 V
It was easy enough with an electron but these are just hard D:
<3 Love you all
a) a proton
b)an alpha particle
The answer is
a)2.0x^-4 V
b)2.57x10^-5 V
It was easy enough with an electron but these are just hard D:
<3 Love you all
-
I can do one, and hopefully, you can do the other.
(a)
w = h/mv
h = planck's constant = 6.626E-34 Js
m = mass = 1.67E-27 kg
v = velocity
2E-9 = 6.626E-34 / 1.67E-27 * v
v = 198 m/s
Use the accelerating potential difference equation:
V = 0.5mv^2 / e
e = 1.6E-19 C
V = 0.5 * 1.67E-27 * 198^2 / 1.6E-19 = 2.05E-4 V
(a)
w = h/mv
h = planck's constant = 6.626E-34 Js
m = mass = 1.67E-27 kg
v = velocity
2E-9 = 6.626E-34 / 1.67E-27 * v
v = 198 m/s
Use the accelerating potential difference equation:
V = 0.5mv^2 / e
e = 1.6E-19 C
V = 0.5 * 1.67E-27 * 198^2 / 1.6E-19 = 2.05E-4 V