Determine the potential (V) by which a proton must be accelerated so as to assume a particle wavelength of 0.0293 nm.
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This is what I did, dunno if right
- wavelength in meters = 2.93*10^-11 m
- used DeBroglie relation to find energy
lambda = h / mv, v from kinetic energy,
lambda = h / sqrt(2*m*E)
2.93*10^-11 = (6.626 * 10^-34) / sqrt(2 * 1.672 *10^-27 * E)
E = 1.529 * 10^-19 J
- E = qV
V = E / q
1.529 * 10^-19 J / 1.602*10^-19 C
V = 0.9544 V
6 answers
That number, if eV, looks ok to me.
thank you for looking into this.
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The result is asked in plain Volts,
it should be tough,because
E is in Joules
q (charge of proton) is in Coulomb
so V = E/q = [Joules]/[Coulomb]
but you're right that's way too big
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The result is asked in plain Volts,
it should be tough,because
E is in Joules
q (charge of proton) is in Coulomb
so V = E/q = [Joules]/[Coulomb]
but you're right that's way too big
@will we're doing same stuff I guess ;)
http://www.rapidtables.com/convert/electric/ev-to-volts.htm
0.9544 is in volts or eV?
volts and eV aren't like "multiples"
eV is an energy, V a potential
taht's V and it's checked correct ;)
eV is an energy, V a potential
taht's V and it's checked correct ;)
0 answers