The de-Broglie wavelength of a neutron having kinetic energy 0.025 eV is . If the kinetic energy of the neutron is reduced to 0.025 eV 8 , its de-Broglie wavelength will be:

Step-by-step solution

The de-Broglie wavelength of a particle with mass m and kinetic energy K is given by the relation = h 2mK . For the initial state, the wavelength is = h 2m(0.025 eV ) . When the kinetic energy is reduced to K' = 0.025 eV 8 , the new wavelength ' is given by ' = h 2m 0.025 eV 8 . Simplifying the expression for ' : ' = h 2m(0.025 eV ) 8 = 8 h 2m(0.025 eV ) . Since 8 = 2 2 , we have ' = 2 2 . Answer: 2 2