Question

What is the wavelength of an electron accelerated through a 30.0-kV potential, as in a TV tube?

a) 0.412 nm
b) 0.624 nm
c) 0.815 nm
d) 1.029 nm

Answer 1

The wavelength of an electron accelerated through a 30.0-kV potential in a TV tube is 0.624 nm.

Step-by-step explanation:

The wavelength of an electron accelerated through a 30.0-kV potential can be calculated using the de Broglie wavelength formula. The formula is given by: λ = h / √(2mE), where λ is the wavelength, h is Planck's constant (6.626 × 10^-34 J s), m is the mass of the electron (9.11 × 10^-31 kg), and E is the energy of the electron (30.0 × 10^3 V).

Plugging in the values, we get: λ = (6.626 × 10^-34 J s) / √(2(9.11 × 10^-31 kg)(30.0 × 10^3 V)) = 0.624 nm.

Therefore, the correct answer is option b) 0.624 nm.