Question

If the mass of neutron = 1.7 x 10-27 kg. then
the De broglie wavelength of neutron of energy
3eV is :
(1) 1.6 x 10-10 m
(2) 1.6 x 10-11 m
(3) 1.4 x 10-10 m
(4) 1.4 x 10-11 m​

Answer 1

Answer:

The De broglie wavelength of neutron is 1.6 × $10^{-11}$   m .

Explanation:

Given as :

The mass of neutron = m = 1.7 × $10^{-27}$  kg

The kinetic energy of neutron = k.E = 3 ev

Since 1 ev = 1.602 × $10^{-19}$ joule

So, 3 ev = 3 × 1.602 × $10^{-19}$ = 4.806 × $10^{-19}$  joule

Let The De-broglie wavelength = $\lambda$

According to question

De-broglie wavelength  = $\dfrac{h}{\sqrt{2\times m\times k.E}}$

where h = Planck constant = 6.63 × $10^{-34}$

So, $\lambda$ = $\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.7\times 10^{-27}\times 4.806\times 10^{-19}}}$

Or,  $\lambda$ = $\dfrac{6.63\times 10^{-34}}{\sqrt{16.3404\times 10^{-46}}}$

Or, $\lambda$ = $\dfrac{6.63\times 10^{-34}}{4.04232\times 10^{-23}}$

Or, $\lambda$ = 1.6 × $10^{-11}$   m

So, The De broglie wavelength =  $\lambda$ = 1.6 × $10^{-11}$   m

Hence, The De broglie wavelength of neutron is 1.6 × $10^{-11}$   m . Answer