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# For What Kinetic Energy of a Neutron Will the Associated De Broglie Wavelength Be 1.40 × 10−10 M? - Physics

For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10−10 m?

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#### Solution

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

Mass of a neutron, mn = 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.

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#### APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 11 Dual Nature of Radiation and Matter
Q 17.1 | Page 408
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