Question

Compare the energy of a photon with that of a neutron when both are associated with wavelength of 1A0, given that the mass of neutron is 1.678 x 10-27 kg.​

Answer 1

Explanation:

sorry I don't know the answer

Answer 2

Answer:

Energy of Photon = 1.9875×10^-15 jule

Energy of Neutron = 1.98746×10^-15 jule

Explanation:

Provided that:

Wavelength (γ) = 1 A°

Mass of neutron (m) = 1.678 x 10^(-27) kg.

We know that Mass of Photon = 0

From Quantam mechanics we know that;

Total energy of a particle :

${E}^{2} = {p}^{2} {c}^{2} + {m0}^{2} {c}^{4} \: \: ...equation(i) \\ or \: {E}^{2} = {h}^{2} {\nu}^{2} + {m0}^{2} {c}^{4} \\ or \: {E}^{2} = {h}^{2} (\frac{ {c}^{2} }{ { \gamma }^{2} } ) + {m0}^{2} {c}^{4} \: \: ...equation(ii)$

For Photon, rest mass (m0) = 0:

Hence Total Energy of Photon;

${E}^{2} = {h}^{2} (\frac{ {c}^{2} }{ { \gamma }^{2} } ) + 0 \\ E = {h} (\frac{ {c}} \gamma ) \\ = \frac{(6.625 \times {10}^{ - 34} \times 3 \times {10}^{8} )}{1 \times {10}^{ - 10} } \: jule \\ = 1.9875 \times {10}^{ - 15} \: jule$

For Neutron, rest mass (m0) = 1.678 x 10^-27 kg

Hence Total Energy of Neutron;

${E}^{2} = {h}^{2} (\frac{ {c}^{2} }{ { \gamma }^{2} } ) + {m0}^{2} {c}^{4} \\ = {(1.9875 \times {10}^{ - 15} )}^{2} + {(1.678 \times 10^-27)}^{2} \times {(3 \times {10}^{ - 8} }^{4} \: jule \\ = 3.95 \times {10}^{ - 30} + 2.280 \times {10}^{ - 84} \: jule \\ E = 1.98746 \times {10}^{ - 15} \: jule$

Hence Total Energy of a Neutron is slightly less than the photon.