Question

Experiments are performed with ultracold neutrons having velocities as small as 1.00 m/s. (a) What is the wavelength of such a neutron? (b) What is its kinetic energy in eV?

a) 6.63 nm, 3.52 x 10^(-14) eV
b) 12.45 nm, 7.81 x 10^(-14) eV
c) 18.29 nm, 1.24 x 10^(-13) eV
d) 25.13 nm, 1.69 x 10^(-13) eV

Answer 1

The wavelength of the neutron is approximately 6.63 nm and its kinetic energy is approximately 3.52 x 10^-14 eV.

Step-by-step explanation:

To find the wavelength of a neutron, we can use the de Broglie wavelength equation: λ = h / p, where λ is the wavelength, h is Planck's constant, and p is the momentum of the neutron. Since momentum is equal to mass times velocity, p = m * v, where m is the mass of the neutron and v is its velocity. Given that the velocity is 1.00 m/s and the mass of a neutron is approximately 1.675×10^-27 kg, we can calculate the wavelength λ.

To find the kinetic energy of the neutron in electron volts (eV), we can use the formula K.E. = 1/2 * m * v^2, where m is the mass of the neutron and v is its velocity. Plugging in the values, we can calculate the kinetic energy of the neutron in eV.

(a) The wavelength of such a neutron is approximately 6.63 nm. (b) Its kinetic energy is approximately 3.52 x 10^-14 eV.