Answer 1

De Broglie wavelength of the neutron, λ = 1.40 × 10⁻¹⁰ m

Mass of a neutron, m_n = 1.66 × 10⁻²⁷ kg

Planck’s constant, h = 6.6 × 10⁻³⁴ Js

Kinetic energy (K) and velocity (v) are related as:

`K = 1/2 m_nv^2` … (1)

De Broglie wavelength (λ) and velocity (v) are related as:

`lambda = h/m_nv` ...(2)

Using equation (2) in equation (1), we get:

`K = 1/2 (m_n h^2)/(lambda^2 m_n^2) = h^2/(2lambda^2m_n)`

`= (6.63 xx 10^(-34))^2/(2xx(1.40 xx  10^(-10))^2 xx 1.66 xx 10^(-27) ) =  6.75 xx 10^(-21) J`

Hence, the kinetic energy of the neutron is 6.75 × 10⁻²¹ J or 4.219 × 10⁻² eV.

Hence, the kinetic energy of the neutron is 6.75 × 10⁻²¹ J or 4.219 × 10−2 eV.