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

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

@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 ;)

To determine the potential (V) required to accelerate a proton to a particle wavelength of 0.0293 nm, you can follow these steps:

1. Convert the given wavelength from nanometers to meters:
- Given wavelength = 0.0293 nm = 0.0293 * 10^-9 m = 2.93 * 10^-11 m

2. Use the de Broglie relation to find the energy of the proton:
λ = h / mv, where λ is the wavelength, h is the Planck's constant (6.626 × 10^-34 J·s), m is the mass of the proton (1.672 × 10^-27 kg), and v is the velocity of the proton.
Rearranging the equation to solve for energy (E):
E = (h^2) / (2 * m * λ^2)

Plugging in the values:
E = (6.626 × 10^-34 J·s)^2 / (2 * 1.672 × 10^-27 kg * (2.93 × 10^-11 m)^2)
E ≈ 1.529 × 10^-19 J

3. Calculate the potential (V) using the energy (E) and the charge of a proton (q):
E = qV, where E is the energy, q is the charge of a proton (1.602 × 10^-19 C), and V is the potential.
Rearranging the equation to solve for V:
V = E / q

Plugging in the values:
V = (1.529 × 10^-19 J) / (1.602 × 10^-19 C)
V ≈ 0.9544 V

Therefore, the potential (V) required to accelerate a proton to a particle wavelength of 0.0293 nm is approximately 0.9544 V.