Question

Calculate the energy in eV corresponding to a wavelength of 1Å for electron and neutron.

Given h=6.6 x 10-34 Js; me= 9.1 x 10-31 kg ; mn= 1.7 x 10-27 kg ; 1 eV = 1.6 x 10-19 J Ans. Ee=

149 eV; En= 0​

Answer 1

Answer:

124.07*10^-18

Explanation:

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Answer 2

The energy of the electron, $E_{e}$ = 149.59 eV

The energy of the neutron, $E_{n}$ = 0.08 eV

Explanation:

Given data,

The planck's constant, h = 6.6 x 10⁻³⁴ Js

De Broglie wavelenth, λ = 1 x 10⁻¹⁰ m

The mass of electron, $m_{e}$ = 9.1 x 10⁻³¹ kg

The mass of neutron, $m_{n}$ = 1.7 x 10⁻²⁷ kg

The De Broglie wavelenth is given by the formula,

                             λ = h / mv

                              v = h / m λ

The K.E is given by,

                             K.E = ½ mv²

                                    = ½ m(h / m λ)²

                                    = ½ h² / m λ²

The K.E of electron

                             K.E = ½$\frac{h^{2} }{m_{e}\lambda^{2}}$

                                    = ½ $\frac{(6.6\times10^{-34})^{2} }{9.1\times10^{-31}(1\times10^{-10})^{2}}$

                                     = 2.393 x 10⁻¹⁷ J

Since 1 eV = 1.6 x 10⁻¹⁹ J

                               $E_{e}$ = 2.393 x 10⁻¹⁷ / 1.6 x 10⁻¹⁹

                                  = 149.59 eV

Hence, the energy of the electron, $E_{e}$ = 149.59 eV

The K.E of neutron,

                              K.E = ½$\frac{h^{2} }{m_{n}\lambda^{2}}$

                                     = ½ $\frac{(6.6\times10^{-34})^{2} }{1.7\times10^{-27}(1\times10^{-10})^{2}}$

                                     = 1.2812 x 10⁻²⁰ J

   ∴                                $E_{n}$ = 1.2812 x 10⁻²⁰  / 1.6 x 10⁻¹⁹

                                      = 0.08 eV

Hence, the energy of the neutron, $E_{n}$ = 0.08 eV