Question

Experiments are performed with ultra cold neutrons having velocities as small as \[{\bf{1}}{\bf{.00}}\;\frac{{\bf{m}}}{{\bf{s}}}\].

(a) What is the wavelength of such a neutron?

(b) What is its kinetic energy in eV?

Short answer

(a) The wavelength of a neutron is obtained as\[3.956 \times {10^{ - 7}}\;{\rm{m}}\].

(b) The kinetic energy in eV is obtained as\[{\rm{5}}{\rm{.23}} \times {\rm{1}}{{\rm{0}}^{ - {\rm{9}}}}{\rm{ eV}}\].

Step-by-step solution

Define formula for the wavelength.

Consider the formula for the proton is:

\[\lambda = \frac{h}{{mv}}\]

Here, h is the plank’s constant, v is the velocity of the photon and m is the mass of the proton.

Determine the expression for kinetic energy as:

\[{{\bf{E}}_{\bf{k}}}{\bf{ = }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{m}}{{\bf{v}}^{\bf{2}}}\]

Evaluating the wavelength

(a)

Substitute the values and solve for the wavelength as:

\[\begin{array}{c}\lambda = \frac{{6.63 \times {{10}^{ - 34}}\;{\rm{J}} \cdot {\rm{s}}}}{{1.67 \times {{10}^{ - 27}}\;{\rm{kg}} \times 1\;\frac{{\rm{m}}}{{\rm{s}}}}}\\ = 3.956 \times {10^{ - 7}}\;{\rm{m}}\end{array}\]

Therefore, the wavelength is\[3.956 \times {10^{ - 7}}\;{\rm{m}}\].

Evaluating the kinetic energy

(b)

Substitute and solve for the kinetic energy as:

\[\begin{array}{c}{E_k} = \frac{1}{2} \times 1.675 \times {10^{ - 27}}\;{\rm{kg}} \times 1\;{{\rm{m}} \mathord{\left/

{\vphantom {{\rm{m}} {\rm{s}}}} \right.

\kern-\nulldelimiterspace} {\rm{s}}}\\ = 8.375 \times {10^{ - 27}}\;{\rm{J}}\\ = 5.23 \times {10^{ - 9}}\;{\rm{eV}}\end{array}\]

Therefore, the kinetic energy\[5.23 \times {10^{ - 9}}\;{\rm{eV}}\].